- Determine the mass of Mg 3 N 2, completely decomposed by water, if 150 ml of a 4% solution was required for salt formation with hydrolysis products of hydrochloric acid density 1.02 g/ml.
Show 1) Mg 3 N 2 + 6H 2 O → 3Mg (OH) 2 + 2NH 3
2) Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O
3) NH 3 + HCl → NH 4 Cl
n (HCl) \u003d 150 * 1.02 * 0.04 / 36.5 \u003d 0.168 mol
Let x mol of Mg 3 N 2 react. According to equation 1, 3x mol of Mg(OH) 2 and 2x mol of NH 3 were formed. Neutralization of 3x mol of Mg(OH) 2 required 6x mol of HCl (according to equation 2), and neutralization of 2x mol of NH 3 required 2x mol of HCl (according to equation 3), for a total of 8x mol of HCl.
8 x \u003d 0.168 mol,
X \u003d 0.021 mol,
n (Mg 3 N 2) \u003d 0.021 mol,
m (Mg 3 N 2) \u003d M * n \u003d 100 * 0.021 \u003d 2.1 g.
Answer: 2.1 g - Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 8.4% sodium bicarbonate solution. What volume of a 15.6% solution of barium chloride (density 1.11 g / ml) will react with the resulting sodium carbonate? Water evaporation can be neglected.
Show 1) 2NaHCO 3 - t → Na 2 CO 3 + H 2 O + CO 2
2) Na 2 CO 3 + BaCl 2 → BaCO 3 + 2 NaCl
n (NaHCO 3) \u003d 150 * 0.084 / 84 \u003d 0.15 mol
From equation (1) n (NaHCO 3) : n (Na 2 CO 3) = 2:1 => n (Na 2 CO 3) = 0.075 mol.
m (Na 2 CO 3) \u003d 0.075 ∙ 106 \u003d 7.95 g
n (CO 2) \u003d 0.075 mol, m (CO 2) \u003d 0.075 ∙ 44 \u003d 3.3 g
m (solution) \u003d 150 - 3.3 \u003d 146.7 g
ω (Na 2 CO 3) \u003d m (Na 2 CO 3) / m (solution) \u003d 7.95 / 146.7 \u003d 0.0542 or 5.42%
From equation (2) n (Na 2 CO 3) : n (BaCl 2) = 1: 1 => n (BaCl 2) = 0.075 mol.
m (BaCl 2) \u003d n * M \u003d 0.075 * 208 \u003d 15.6 g.
m (solution) \u003d m (BaCl 2) / ω \u003d 15.6 / 0.156 \u003d 100 g
V (solution) \u003d m (solution) / ρ \u003d 100 / 1.11 \u003d 90.1 ml.
Answer: 5.42%, 90.1 ml. - In what mass ratios should 10% sodium hydroxide and sulfuric acid solutions be mixed to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in this solution?
Show 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
Let the mass of NaOH solution be 100 g, m(NaOH)= 10 g,
n(NaOH)=0.25 mol,
n(H 2 SO 4) = 0.125 mol,
m (H 2 SO 4) \u003d 12.25 g,
m (H 2 SO 4 solution) = 122.5 g
Ratio m (NaOH solution): m (H 2 SO 4 solution) = 1:1.2
n(Na 2 SO 4) = 0.125 mol,
m (Na 2 SO 4) \u003d 17.75 g,
m (solution) \u003d 100 + 122.5 g \u003d 222.5 g,
w(Na 2 SO 4)=7.98% - How many liters of chlorine (N.O.) will be released if 26.1 g of manganese (IV) oxide are added to 200 ml of 35% hydrochloric acid (density 1.17 g / ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?
Show 1) MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2
2) 2NaOH + Cl 2 → NaCl + NaClO + H 2 O
n (HCl) \u003d 200 * 1.17 * 0.35 / 36.5 \u003d 2.24 mol - in excess
n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 mol - in short supply
According to equation (1) n (Cl 2) \u003d 0.3 mol
V (Cl 2) \u003d 6.72 l
According to equation (2) n(NaOH)= 0.6 mol, m(NaOH)=24 g
Answer: 6.72 l, 24 g - In what volume of water should 11.2 liters of sulfur oxide (IV) (n.a.) be dissolved in order to obtain a solution of sulfurous acid with a mass fraction of 1%? What color will litmus acquire when it is added to the resulting solution?
- What mass of lithium hydride must be dissolved in 100 ml of water to obtain a solution with a mass fraction of hydroxide of 5%? What color will litmus acquire when it is added to the resulting solution?
Show LiH + H 2 O → LiOH + H 2
Let m(LiH)= x g, then m (LiOH)=x*24/8 = 3x g.
m (solution) \u003d m (H 2 O) + m (LiH) - m (H 2)
m (r-ra) \u003d x + 100 - x / 4 \u003d 0.75x + 100
w \u003d m (in-va) * 100% / m (r-ra)
3x/(0.75x+100) = 0.05
3x=0.038x+5
2.96x = 5
x=1.7 g - In what mass of a solution with a mass fraction of Na 2 SO 4 10% should 200 g of Na 2 SO 4 × 10H 2 O be dissolved in order to obtain a solution with a mass fraction of sodium sulfate 16%? What medium will the resulting solution have?
Show Let the mass of the solution be x g. It contains 0.1 x g of Na 2 SO 4.
200 g of crystalline hydrate was added, in which the mass of sodium sulfate is 200*142/322=88.2 g.
(0.1x +88.2) / (x + 200) \u003d 0.16
0.1x +88.2 = 0.16x + 32
0.06x = 56.2
x = 937
Answer: 937 g, neutral. - Gaseous ammonia released by boiling 160 g of a 7% potassium hydroxide solution with 9.0 g of ammonium chloride was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
- Ammonia released by boiling 80 g of a 14% potassium hydroxide solution with 8.03 g of ammonium chloride was dissolved in water. Calculate how many milliliters of 5% nitric acid with a density of 1.02 g / ml will be used to neutralize the resulting ammonia solution.
Show KOH + NH 4 Cl → NH 3 + H 2 O + KCl
NH 3 + HNO 3 → NH 4 NO 3
n (NH 4 Cl) \u003d 8.03 / 53.5 \u003d 0.15 mol
m (KOH) \u003d 80 * 0.14 \u003d 11.2 g
n (KOH) \u003d 11.2 / 56 \u003d 0.2 mol
KOH is in excess.
Further calculations are carried out according to the deficiency
n (NH 4 Cl) \u003d n (NH 3) \u003d 0.15 mol
n (HNO 3) \u003d 0.15 mol
m (HNO 3) \u003d n * M \u003d 0.15 * 63 \u003d 9.45 g
m (p-ra HNO 3) \u003d m (in-va) * 100% / w \u003d 9.45 / 0.05 \u003d 189 g
m = V*ρ
V= m / ρ= 189 / 1.02 = 185.3 ml
Answer: 185.3 ml - Calcium carbide treated with excess water. The released gas occupied a volume of 4.48 liters (N.O.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g / ml will be used to completely neutralize the alkali formed from calcium carbide.
Show 1) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
n (C 2 H 2) \u003d V / V m \u003d 4.48 / 22.4 \u003d 0.2 mol
From equation (1) => n(C 2 H 2) = n(Ca(OH) 2)
n (Ca (OH) 2) \u003d 0.2 mol
2) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O
From equation (2) => n(Ca(OH) 2) : n(HCl) = 1:2 => n(HCl) = 0.4 mol
m(HCl) \u003d n * M \u003d 0.4 * 36.5 \u003d 14.6 g
m (HCl solution) \u003d 14.6 / 0.2 \u003d 73 g
V(HCl solution) = 73/1.1 = 66.4 ml - Calculate what volume of a 10% hydrogen chloride solution with a density of 1.05 g/ml will be used to completely neutralize the calcium hydroxide formed during the hydrolysis of calcium carbide, if the gas released during hydrolysis occupies a volume of 8.96 l (n.o.).
Show CaC 2 + 2 H 2 O → Ca (OH) 2 + C 2 H 2
Ca(OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O
n(C 2 H 2)= V / V n.o.s. = 8.96 / 22.4 = 0.4 mol
n (C 2 H 2) \u003d n Ca (OH) 2 \u003d 0.4 mol
n(HCl) = 0.4 * 2 = 0.8 mol
m (in HCl) \u003d n * M \u003d 0.8 * 36.5 \u003d 29.2 g
w \u003d m (in-va) * 100% / m (r-ra)
m (solution) \u003d m (in-va) * 100% / w \u003d 29.2 / 0.1 \u003d 292 g
m = V*ρ
V= m / ρ = 292 / 1.05 = 278ml
Answer: 278 ml - Aluminum carbide treated with 200 g of a 30% sulfuric acid solution. The methane released at the same time occupied a volume of 4.48 l (n.o.). Calculate the mass fraction of sulfuric acid in the resulting solution.
Show Al 4 C 3 + 6H 2 SO 4 → 2Al 2 (SO 4) 3 + 3CH 4
n(CH 4) \u003d V / Vm \u003d 4.48 / 22.4 \u003d 0.2 mol
m (CH 4) \u003d m * M \u003d 0.2 * 16 \u003d 3.2 g
m(Al 4 C 3) \u003d 1/3 * 0.2 * 144 \u003d 9.6 g
Reacted, according to the equation n (H 2 SO 4) \u003d 0.4 mol,
m (H 2 SO 4) \u003d 0.4 * 98 \u003d 39.2 g.
Initially, m (H 2 SO 4) = m (p-ra) * ω = 200 g * 0.3 = 60 g was added. m (H 2 SO 4) = 60 - 39.2 = 20.8 g remained.
m (H 2 SO 4) \u003d 0.21 * 98 \u003d 20.8 g
m (p-pa) \u003d m (Al 4 C 3) + m (p-pa H 2 SO 4) - m (CH 4)
m(p-pa) = 9.6 g + 200 g - 3.2 g = 206.4 g
ω (H 2 SO 4) \u003d m (H 2 SO 4) / m (solution) \u003d 20.8 / 206.4 * 100% \u003d 10% - When processing aluminum carbide with a solution of hydrochloric acid, the mass of which is 320 g and the mass fraction of HCl 22%, 6.72 l (n.o.) of methane were released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
Show Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 n (CH 4) \u003d V / Vm \u003d b.72 / 22.4 \u003d 0.3 mol;
According to the equation n (HCl) \u003d 4 n (CH 4) \u003d 1.2 mol
m (HCl) \u003d m (solution) * ω \u003d 320 0.22 \u003d 70.4 g;
Reacted m (HCl) \u003d 1.2 36.5 \u003d 43.8 g.
Remaining m(HCl) = 70.4 - 43.8 = 26.6 g.
m (p-pa) \u003d 320 g + m (Al 4 C 3) - m (CH 4),
According to the equation n (Al 4 C 3) \u003d 1/3 n (CH 4) \u003d 0.1 mol;
m (Al 4 C 3) \u003d 0.1 144 \u003d 14.4 g,
m(CH 4) \u003d 0.3 16 \u003d 4.8 g,
m(p-pa) \u003d 320 g + 14.4 g - 4.8 g \u003d 329.6 g.
ω(HCl) = 26.6 / 329.6 100% = 8.07% - Calcium hydride was added to an excess of hydrochloric acid solution (mass of acid solution 150 g, mass fraction of HCl 20%). In this case, 6.72 l (N.O.) of hydrogen were released. Calculate the mass fraction of calcium chloride in the resulting solution.
Show n (CaCl 2) \u003d ½ n (H 2) \u003d 0.15 mol
m (CaCl 2) \u003d 111 * 0.15 \u003d 16.65 g
W (CaCl 2) \u003d m in-va / m solution \u003d 16.65 / 155.7 \u003d 0.1069 or 10.69%
Answer: W (CaCl 2) \u003d 10.69% - 125 ml of 5% lithium hydroxide solution (r = 1.05 g/ml) and 100 ml of 5% nitric acid solution (ρ = 1.03 g/ml) were mixed. Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
Show LiOH + HNO₃ = LiNO₃ + H₂O
m (LiOH solution)= V × ρ = 125 ml × 1.05 g/ml = 131.25 g
m(LiOH) = 131.25 g × 0.05 = 6.563 g
n(LiOH) \u003d m / M \u003d 6.563 / 24 \u003d 0.273 mol
m (solution HNO₃:) = V × ρ = 100 ml × 1.03 g/ml = 103 g
m(HNO₃) = 103 g × 0.05 = 5.15 g
n(HNO₃) = 5.15 / 63 = 0.0817 mol
LiOH is given in excess, the calculation is based on acid.
n(LiNO₃) = 0.0817 mol
m(LiNO₃) = n × M = 0.0817 × 69 = 5.64 g
m (solution obtained) \u003d m (LiOH solution) + m (HNO₃ solution) \u003d 131.25 g + 103 g \u003d 234.25 g
ω(LiNO₃) = 5.64 / 234.25 × 100% = 2.4%
Answer: alkaline, 2.4%; - Phosphorus (V) oxide weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid, and the resulting solution was boiled. What salt and in what quantity is formed if 3.92 g of potassium hydroxide is added to the resulting solution?
- Sulfur oxide (VI) weighing 8 g was dissolved in 110 g of 8% sulfuric acid. What salt and in what quantity is formed if 10.6 g of potassium hydroxide is added to the resulting solution?
Show SO 3 + H 2 O \u003d H 2 SO 4
n \u003d m / MM (SO 3) \u003d 80 g / mol,
n(SO 3) \u003d 8/80 \u003d 0.1 mol.
According to equation (1) n (H 2 SO 4) \u003d n (SO 3) \u003d 0.1 mol,
n(KOH)=10.6/56=0.19 mol.
In the original solution n (H 2 SO 4) \u003d 110 * 0.08 / 98 \u003d 0.09 mol.
After adding sulfur oxide n (H 2 SO 4) \u003d 0.09 + 0.1 \u003d 0.19 mol.
The ratio of alkali and acid is 1:1, which means that an acid salt is formed.
H 2 SO 4 + KOH \u003d KHSO 4 + H 2 O
n (H 2 SO 4) \u003d n (KOH) \u003d n (KHSO 4) \u003d 0.19 mol
Answer: KHSO 4, 0.19 mol. - Ammonia released during the interaction of 107 g of 20% ammonium chloride solution with 150 g of 18% sodium hydroxide solution completely reacted with 60% phosphoric acid to form ammonium dihydrogen phosphate. Determine the mass fraction of sodium chloride in the solution and the required mass of a 60% phosphoric acid solution.
Show NH 4 Cl + NaOH = NaCl + NH 3 + H 2 O
m(NH 4 Cl) \u003d 107 g ∙ 0.2 \u003d 21.4 g
n(NH 4 Cl) \u003d 21.4 g / 53.5 g / mol \u003d 0.4 mol
m(NaOH) = 150 g ∙ 0.18 = 27 g
n(NaOH) = 27 g / 40 g / mol = 0.675 mol, therefore, NaOH is in excess
n(NaCl) = n(NH4Cl) = 0.4 mol
m(NaCl) = 0.4 ∙ 58.5 = 23.4 g
n (NH 3) \u003d n (NH 4 Cl) \u003d 0.4 mol
m (NH 3) \u003d 0.4 ∙ 17 \u003d 6.8 g
m (solution) \u003d m (solution NH 4 Cl) + m (solution NaOH) - m (NH 3) \u003d 107 + 150 - 6.8 \u003d 250.2 g
w(NaCl) = 23.4 / 250.2 = 0.094 or 9.4%
NH 3 + H 3 PO 4 \u003d NH 4 H 2 PO 4
n (NH 3) \u003d n (H 3 PO 4) \u003d 0.4 mol
m (H 3 PO 4) \u003d 98 ∙ 0.4 \u003d 39.2 g
m (solution H 3 PO 4) \u003d 39.2 / 0.6 \u003d 65.3 g - Hydrogen sulfide, released during the interaction of an excess of concentrated sulfuric acid with 1.44 g of magnesium, was passed through 160 g of a 1.5% bromine solution. Determine the mass of the precipitate formed in this case and the mass fraction of acid in the resulting solution.
Show 4Mg + 5H 2 SO 4 \u003d 4MgSO 4 + H 2 S + 4H 2 O
H 2 S + Br 2 \u003d 2HBr + S ↓
n(Mg) \u003d m / M \u003d 1.44 g: 24 g / mol \u003d 0.06 mol
n(H 2 S) = ¼ n(Mg) = 0.015 mol
m(H 2 S) \u003d n * M \u003d 0.015 mol * 34 g / mol \u003d 0.51 mol
m (in-va Br 2) \u003d 160 g * 0.015 \u003d 2.4 g
n(Br 2) \u003d m / M \u003d 2.4 g: 160 g / mol \u003d 0.015 mol
n (HBr) \u003d 2n (Br 2) \u003d 0.03 mol
m(HBr) \u003d n * M \u003d 0.03 mol * 81 g / mol \u003d 2.43 g
n(S) \u003d n (Br 2) \u003d 0.015 mol
m(S) = n * M = 0.015 mol * 32 g/mol = 0.48 g
m (solution) \u003d m (H 2 S) + m (solution Br 2) -m (S) \u003d 0.51 g + 160 g - 0.48 \u003d 160.03 g
W (HBr) \u003d m (HBr) / m (solution) \u003d 2.43 g / 160.03 g \u003d 0.015 or 1.5%
Answer: m (S) \u003d 0.48 g, w (HBr) \u003d 1.5% - Chlorine reacted without residue with 228.58 ml of 5% NaOH solution (density 1.05 g/ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
Show 6NaOH + 3Cl 2 = 5NaCl + NaClO 3 + 3H 2 O (at t)
m solution \u003d 228.58 ∙ 1.05 \u003d 240 g;
m(NaOH) \u003d 240 ∙ 0.05 \u003d 12g.
n(NaOH) = 12/40 = 0.3 mol;
n(Cl 2) = 0.15 mol;
n(NaCl) = 0.25 mol;
n(NaClO 3) = 0.05 mol
m (NaCl) \u003d 58.5 ∙ 0.25 \u003d 14.625 g;
m (NaClO 3) \u003d 106.5 ∙ 0.05 \u003d 5.325 g:
m solution \u003d 240 + m (Cl 2) \u003d 240 + 71 ∙ 0.15 \u003d 240 + 10.65 \u003d 250.65 g
W(NaCl) = 14.625 / 250.65 = 0.0583 or 5.83%
W (NaClO 3) \u003d 5.325 / 250.65 \u003d 0.0212 or 2.12% - Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ = 1.153 g/ml). For complete binding of the products, 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the used solution.
Show 3Cu + 8HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O
m (HNO 3) \u003d 100 ∙ 0.3 ∙ 1.153 \u003d 34.59g
n(HNO 3) = 34.59/ 63 = 0.55 mol, n(Cu) = 6.4/ 64 = 0.1 mol
n (HNO 3) g = 0.55 - 8/3 ∙ 0.1 = 0.28 mol
Cu (NO 3) 2 + 2 NaOH \u003d Cu (OH) 2 + 2 NaNO 3
HNO 3 + NaOH \u003d NaNO 3 + H 2 O
n(NaOH) \u003d n (HNO 3) g + 2n (Cu (NO 3) 2) \u003d 0.28 + 0.1 ∙ 2 \u003d 0.48 mol
m (NaOH) \u003d 0.48 ∙ 40 \u003d 19.2 g.
W(NaOH) = 19.2/ 200 = 0.096 or 9.6% - In 60 g of 18% orthophosphoric acid, 2.84 g of phosphorus (V) oxide was dissolved and the resulting solution was boiled. What salt and in what quantity is formed if 30 g of sodium hydroxide is added to the resulting solution?
Show 1) 3H 2 O + P 2 O 5 → 2H 3 PO 4
In the original solution m (H 3 PO 4) \u003d m (solution) * ω \u003d 60 * 0.18 \u003d 10.8 g.
n(P 2 O 5) = m/M = 2.84/142 = 0.02 mol
As a result of the reaction, m (H 3 PO 4) \u003d 0.04 * 98 \u003d 3.92 g was formed
Total m (H 3 PO 4) \u003d 3.92 + 10.8 \u003d 14.72 g.
n (H 3 PO 4) \u003d m / M \u003d 14.72 / 98 \u003d 0.15 mol
n(NaOH) \u003d m / M \u003d 30/40 \u003d 0.75 mol - in excess, salt is average.
2) 3NaOH + H 3 PO 4 → Na 3 PO 4 + 3H 2 O
According to equation (2) n (Na 3 PO 4) \u003d n (H 3 PO 4) \u003d 0.15 mol
m (Na 3 PO 4) \u003d 0.15 * 164 \u003d 24.6g
Answer: 24.6 g - Ammonia with a volume of 4.48 l (n.o.) was passed through 200 g of a 4.9% solution of phosphoric acid. Name the salt formed as a result of the reaction, and determine its mass.
- 5.6 l (n.o.) of hydrogen sulfide reacted without residue with 59.02 ml of 20% KOH solution (density 1.186 g/ml). Determine the mass of the salt produced by this chemical reaction.
Show m (p-ra KOH) \u003d V * ρ \u003d 1.186 * 59.02 \u003d 70 g,
m (KOH) \u003d m (p-raKOH) * ω \u003d 70 g * 0.2 \u003d 14 g,
n(KOH) \u003d m / M \u003d 14/56 \u003d 0.25 mol,
n(H 2 S) \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.
The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acid salt is formed - hydrosulfide according to the reaction: H 2 S + KOH \u003d KHS + H 2 O
According to the equation n(KHS) = 0.25 mol,
m(KHS) \u003d M * n \u003d 72 0.25 \u003d 18 g.
Answer: 18 - To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution (density 1.20 g/ml) was used. Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
Show COOH + KOH \u003d NSOOK + H 2 O
CH 3 COOH + KOH \u003d CH 3 COOK + H 2 O
m (p-paKOH) \u003d V (p-pa) * ρ \u003d 35 * 1.2 \u003d 42 g.
m(KOH) \u003d m (p-pa) * ω (KOH) \u003d 42 * 0.2 \u003d 8.4 g
n(KOH) \u003d m (KOH) / M (KOH) \u003d 8.4 / 56 \u003d 0.15 mol
Let n (HCOOH) \u003d x mol, and n (CH 3 COOH) \u003d y mol.
m (HCOOH) \u003d n (HCOOH) * M (HCOOH) \u003d x * 46 g
m (CH 3 COOH) \u003d n (CH 3 COOH) * M (CH 3 COOH) \u003d y * 60 g
Let's make a system of equations:
x + y = 0.15
60y + 46x = 7.6
We solve the system: x \u003d 0.1 mol, y \u003d 0.05 mol
m (CH 3 COOH) \u003d n (CH 3 COOH) * M (CH 3 COOH) \u003d 0.05 * 60 \u003d 3 g.
ω(CH 3 COOH) = m(CH 3 COOH)/m(mixture) = 3/7.6 = 0.395 or 39.5%.
Answer: 39.5% - 100 ml of 30% perchloric acid solution (r = 1.11 g/ml) and 300 ml of 20% sodium hydroxide solution (r = 1.10 g/ml) were mixed. How many milliliters of water should be added to the resulting mixture so that the mass fraction of sodium perchlorate in it would be 8%?
Show HClO 4 + NaOH = NaClO 4 + H 2 O
m (p-p NaOH) \u003d V * ρ \u003d 300 * 1.10 \u003d 330 g.
n (NaOH) \u003d m (p-p NaOH) * ω / M \u003d 330 * 0.2 / 40 \u003d 1.65 mol - in excess.
m (solution HClO 4) \u003d V * ρ \u003d 100 * 1.11 \u003d 111 g.
n (HClO 4) \u003d 111 * 0.3 / 100.5 \u003d 0.331 mol,
According to the equation n (HClO 4) \u003d n (NaClO 4) \u003d 0.331 mol,
m (NaClO 4) \u003d n * M \u003d 0.331 * 122.5 \u003d 40.5 g.
Let the mass of added water be x g.
40.5 / (111 + 330 + x) \u003d 0.08
whence x = 65.3 g.
V(H 2 O)=65.3 ml.
Answer: 65.3 ml - To 100 ml of a 5% hydrochloric acid solution (density 1.02 g/ml) was added 6.4 g of calcium carbide. How many milliliters of 15% nitric acid (density 1.08 g/ml) should be added to the resulting mixture to completely neutralize it?
Show 1) CaC 2 + 2HCl \u003d CaCl 2 + C 2 H 2
2) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
3) Ca(OH) 2 + 2HNO 3 = Ca(NO 3) 2 + 2H 2 O
n(HCl) \u003d m (HCl) / M (HCl) \u003d Vp-ra (HCl) * (HCl) * p (HCl) / M (HCl) \u003d 100 * 0.05 * 1.02 / 36.5 \u003d 0.14 mol,
n (CaC 2) \u003d m / M \u003d 6.4 / 64 \u003d 0.1 mol.
According to equation (1) n (CaC 2) : n (HCl) = 1: 2 => CaC 2 - in excess.
Reacted n (CaC 2) = n (HCl) / 2 = 0.07 mol.
It remains n (CaC 2) \u003d 0.1 - 0.07 \u003d 0.03 mol.
According to equation (2) n (CaC 2) \u003d n (Ca (OH) 2) \u003d 0.03 mol.
By equation (3) n(Ca(OH)2) : n(HNO3) = 1: 2 =>
n (HNO 3) \u003d 2n (Ca (OH) 2) \u003d 0.06 mol.
Vr-ra (HNO 3) \u003d m (r-ra) / ρ
m (solution) \u003d m (HNO 3) / ω \u003d 0.06 * 63 / 0.15 \u003d 25.2 g,
V (p-raHNO 3) \u003d 25.2 / 1.08 \u003d 23.3 ml.
Answer: 23.3 ml - Sodium nitrite weighing 13.8 g was introduced by heating into 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution?
Show NaNO 2 + NH 4 Cl \u003d N 2 + NaCl + 2H 2 O
n (NaNO 2) \u003d 13.8 / 69 \u003d 0.2 mol
n (NH 4 Cl) \u003d 220 0.1 / 53.5 \u003d 0.41 mol
NH 4 Cl - in excess n (N 2) \u003d n (NaNO 2) \u003d 0.2 mol
V (N 2) \u003d 0.2 mol 22.4 l / mol \u003d 4.48 l
Calculate the mass of ammonium chloride remaining in excess:
n(NH 4 Cl) g = 0.41 - 0.2 = 0.21 mol
m (NH 4 Cl) g = 0.21 53.5 = 11.2 g Calculate the mass fraction of ammonium chloride:
m(p-pa) \u003d 13.8 + 220 - 0.2 28 \u003d 228.2 g
ω (NH 4 Cl) \u003d 11.2 / 228.2 \u003d 0.049 or 4.9% Answer:
V (N 2) \u003d 4.48 l
ω(NH 4 Cl) \u003d 4.9% - Potassium nitrite weighing 8.5 g was introduced by heating into 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution?
Show KNO 2 + NH 4 Br \u003d N 2 + KBr + 2H 2 O
m (in-va NH 4 Br) \u003d 270 g * 0.12 \u003d 32.4 g
n(NH 4 Br) \u003d m / M \u003d 32.4 g: 98 g / mol \u003d 0.33 mol
n(KNO 2) \u003d m / M \u003d 8.5 g: 85 g / mol \u003d 0.1 mol
n(NH 4), proreact. with KNO 2 \u003d 0.33 mol - 0.1 mol \u003d 0.23 mol (since n (KNO 2) : n (NH 4 Br) \u003d 1: 1)
m (NH 4 Br) remaining in the final solution \u003d n * M \u003d 0.23 mol * 98 g / mol \u003d 22.54 g
n (N 2) \u003d n (KNO 2) \u003d 0.1 mol
(N 2) \u003d n * M \u003d 0.1 mol * 28 g / mol \u003d 2.8 g
V (N 2) \u003d n * M \u003d 0.1 mol * 22.4 l / mol \u003d 2.24 l
m (final solution) \u003d m (KNO 2) + m (NH 4 Br solution) - m (N 2) \u003d 8.5 g + 270 g - 2.8 g \u003d 275.7 g
W(NH 4 Br in final solution) = 22.54 g: 275.7 g = 8%
Answer: V (N 2) \u003d 2.24 l; W(NH 4 Br) = 8% - Mixed 300 ml of a solution of sulfuric acid with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of a solution of potassium hydroxide with a mass fraction of 20% (density of 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%?
- In 120 ml of a solution of nitric acid with a mass fraction of 7% (density 1.03 g/ml) made 12.8 g of calcium carbide. How many milliliters of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it?
Show 1) CaC 2 + 2HNO 3 \u003d Ca (NO 3) 2 + C 2 H 2.
n (CaC 2) \u003d 12.8 / 64 \u003d 0.2 mol
n (HNO 3) \u003d (0.07 1.03 120) / 63 \u003d 0.137 mol
CaC 2 - excess. 2) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2.
n (Ca (OH) 2) \u003d 0.2 - 0.137 / 2 \u003d 0.13 mol 3) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O.
n(HCl) \u003d 0.13 2 \u003d 0.26 mol
m (solution) \u003d m (HCl) / w \u003d (0.26 36.5) / 0.2 \u003d 47.45 g
V (solution HCl) \u003d m (solution) / ρ \u003d 47.45 / 1.10 \u003d 43.1 ml.
Answer: 43.1 ml. - To the solution obtained by adding 4 g of potassium hydride to 100 ml of water was added 100 ml of a 39% solution of nitric acid (r = 1.24 g/ml). Determine the mass fractions of all substances (including water) in the final solution.
Show KH + H 2 O \u003d KOH + H 2
n(KH) = 4 g: 40 g/mol = 0.1 mol
n(H 2 O) \u003d 100 g: 18 g / mol \u003d 5.6 mol
KH is in short supply, then n (KOH) \u003d n (KH) \u003d 0.1 molKOH + H 2 NO 3 \u003d KNO 3 + H 2 O
a) m(KOH) = 0.1 * 56 = 5.6
b) m in-va (HNO 2) \u003d 100 ml * 1.24 g / ml * 0.39 \u003d 48.36 g
n (HNO 3) \u003d 48.36 g: 63 g / mol \u003d 0.77 mol
HNO 3 in excess, n excess (HNO 3) \u003d 0.77 - 0.1 \u003d 0.67 mol
m (HNO 3) \u003d 0.67 * 63 \u003d 42.21 g
m (solution) \u003d 4g + 100g + 124g - 0.2g \u003d 227.8g
3) W (KNO 3) \u003d m (KNO 3) : m (p-pa) \u003d (0.1 mol * 101 g / mol) 227.8 g * 100% \u003d 4.4%
W (HNO 3) \u003d 42.21: 227.8 * 100% \u003d 18.5
W (H 2 O) \u003d 100% - (W (KNO 3) + W (HNO 3)) \u003d 77.1%
Answer:
W(KNO 3) = 4.4%
W(HNO 3) \u003d 18.5%
W(H 2 O) = 77.1%1) 2Na 2 O 2 + 2H 2 O \u003d 4NaOH + O 2
2) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
m (solution H 2 SO 4) \u003d 300 * 1.08 \u003d 324 g
m (H 2 SO 4) \u003d 0.1 * 324 \u003d 32.4 g
n (H 2 SO 4) \u003d 32.4 / 98 \u003d 0.33 mol
n(NaOH): n(H 2 SO 4) = 2:1 => n(NaOH) = 0.33 * 2 = 0.66 mol
n(Na 2 O 2): n(NaOH) = 1:2 => n(Na 2 O 2) = 0.66/2 = 0.33 mol
m (Na 2 O 2) \u003d n * M \u003d 0.33 * 78 \u003d 25.7 g
n(Na 2 O 2): n(O 2) = 2:1 => n(O 2) = 0.33/2 = 0.165 mol
V (O 2) \u003d 0.165 * 22.4 \u003d 3.7 l
Answer: m (Na 2 O 2) \u003d 25.7 g; V (O 2) \u003d 3.7 l - When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the initial solution, by heating which you can get an 8% solution of potassium carbonate.
- When interacting in a sulfuric acid medium, 17.4 g of manganese dioxide with 58 g of potassium bromide, bromine was released at a 77% yield. What volume (N.O.) of propene can react with the resulting amount of bromine?
- Carbon dioxide with a volume of 5.6 l (n.o.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ = 1.22 g/ml). Determine the composition and mass fractions of substances in the resulting solution.
- Aluminum carbide was dissolved in a 15% sulfuric acid solution weighing 300 g. The methane released at the same time occupied a volume of 2.24 l (n.o.). Calculate the mass fraction of sulfuric acid in the resulting solution.
- In an excess of oxygen, 8 g of sulfur were burned. The resulting gas was passed through 200 g of 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
- A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 L (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.
- A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
2KHCO 3 + H 2 SO 4 = K 2 SO 4 + 2CO 2 + 2H 2 O (2) Na 2 CO 3 + 2HCl \u003d NaCl + CO 2 + H 2 O
NaHCO 3 + HCl \u003d NaCl + CO 2 + H 2 O
NaHCO 3 + NaOH \u003d Na 2 CO 3 + H 2 O
1) findamountNaHCO3
n(NaOH)=(NaHCO 3)=m(NaOH) . ω (NaOH) / M (NaOH) \u003d 80 * 0.1 / 40 \u003d 0.2 mol
2) findmassNaHCO3
m(NaHCO 3)=n(NaHCO 3)*M(NaHCO 3)=0.2. 84=16.8 g
3) find the amount of Na 2 CO 3
n(HCl) = m(HCl) . ω (HCl) / M (HCl) \u003d 73 * 0.2 / 36.5 \u003d 0.4 mol
n (Na 2 CO 3) \u003d (n (HCl) - n (NaOH)) / 2 \u003d 0.1 mol
4) find the massNa 2 CO 3
m(Na 2 CO 3)=n. M = 0.1. 106 = 10.6 g
5) find the mass fractionNa2CO3
ω (Na 2 CO 3) \u003d m (Na 2 CO 3) / m (mixture) . 100%=38.7%
m(mixture).=m(Na 2 CO 3)+m(NaHCO 3)=27.4 g
Answer: ω (Na 2 CO 3) \u003d 38.7%1) 2Na + 2H 2 O → 2NaOH + H 2
2) Na 2 O + H 2 O → 2NaOH
1) find the quantityNa
n(H 2) \u003d V / Vm \u003d 4.48 / 22.4 \u003d 0.2 mol
According to equation (1) n (Na) \u003d n 1 (NaOH) \u003d 2. n(H 2) = 0.4 mol
2) find the massNa
m(Na) = n . M = 0.4. 23 = 9.2 g
3) find the quantityNaOH
n(NaOH) = m(p-p) . ω/M(NaOH) = 240 . 0.1/40 = 0.6 mol
4) find the quantityNa 2 O
n (Na 2 O) \u003d (n (NaOH) - n 1 (NaOH)) / 2 \u003d 0.1 mol
m(Na 2 O)= n. M = 0.1. 62 = 6.2g
5) findmmass shareNa
ω (Na) = m(Na)/m(mixtures) . 100%=59.7%
m(mixture)=m(Na)+m(Na 2 O)=15.4
Answer: ω (Na)=59,7%
1) Find the totalH2SO4
n 1 (H 2 SO 4) \u003d m. ω/M=490 . 0.4/98=2 mol
M(H 2 SO 4) \u003d 2 + 32 + 64 \u003d 98 g / mol
2) Let's findH2SO4reactedwithNa2CO3
n (Na 2 CO 3) \u003d n (Na 2 CO 3. 10H 2 O) \u003d m / M \u003d 143/286 \u003d 0.5 mol
M(Na 2 CO 3. 10H 2 O) \u003d (46 + 12 + 48) + (10. 18) \u003d 286 g / mol
According to the equation n (H 2 SO 4) \u003d n (Na 2 CO 3) \u003d 0.5 mol
3) findH 2 SO 4 reacted withNaOH
n (H 2 SO 4) \u003d n 1 -n \u003d 2-0.5 \u003d 1.5 mol
4) find the mass of NaOH
n(NaOH)=2n(H 2 SO 4)=2. 1.5=3 mol
m(NaOH)=n. M=3 . 40=120 g
M(NaOH)=23+16+1=18 g/mol
5) find the mass fractionNaOH
ω (NaOH) \u003d m (substance) / . m(p-p)*100%=120/1200 . 100%=10%
Answer: m(NaOH)=120g; ω(NaOH)=10%
48) A mixture of magnesium and aluminum filings was treated with an excess of dilute hydrochloric acid, and 11.2 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of potassium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of hydrobromic acid (ρ = 1.12 g / ml) with a mass fraction of 20%. What is the mass fraction of hydrogen bromide in the resulting solution?
Types of tasks Tasks on a mixture The task condition contains the words: “mixture”, “technical”, “impurity”, names of minerals or alloys Tasks on solutions The task condition contains the words: “solution”, “mass fraction of a dissolved substance” Tasks on excess Disadvantage The task condition contains information about both reagents Product Yield Tasks The task condition contains the words: “substance yield”, “product yield mass fraction” Tasks in which the products of one reaction are used to carry out another reaction.
Here's the challenge... Reacting 6.3 grams of a mixture of aluminum and magnesium with sulfuric acid requires 275.8 ml of a 10% sulfuric acid solution (density 1.066 g/ml). Determine what mass of a 20% solution of barium chloride will be required for the complete precipitation of metal sulfates from the resulting solution.
Let's determine the amount of sulfuric acid substance in the solution: m (p-ra) = V (p-ra). = 278.8. 1.066 \u003d 294 (g) m (H 2 SO 4) \u003d m (solution). w (H 2 SO 4) \u003d, 1 \u003d 29.4 g n (H 2 SO 4) \u003d 29.4 / 98 \u003d 0.3 mol Another 1 point!
Let's determine the mass of the solution: n (BaCl2) = 0.3 mol m (BaCl2) = .3 = 62.4 (g) m (p-ra BaCl2) = 62.4: 0.2 = 312 (g) Answer: m (p-ra BaCl 2) \u003d 312 g And 1 more point Total - 4 points Total - 4 points
Elements of solving the problem and evaluating the results 1. The equations of all reactions are written correctly. 2. The “amount of substance” (volume, mass) of the initial substances was found. 3. A system of equations was compiled to find the amount of substances in the mixture. 4. The final data, which is asked in the problem, has been calculated. All elements are correctly performed - 4 points One mistake is made - 3 points Two mistakes are made - 2 points Three mistakes are made - 1 point All elements are performed incorrectly - 0 points
Problems for self-solving A mixture of zinc and zinc carbonate was treated with an excess of hydrochloric acid solution, and 13.44 liters of gas (n.a.) were released. The gas was burned, the combustion products were cooled to the previous temperature, while the volume of gas decreased to 8.96 liters. What was the composition of the original mixture of substances?
Homework 1 Sodium nitrite weighing 13.8 g was added by heating to 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution? Answer: w (NH 4 Cl) \u003d 4.9%
Homework 2 Potassium nitrite weighing 8.5 g was added by heating to 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution? Answer: V (N 2) \u003d 2.24 l, w (NH 4 Br) \u003d 8.2%
Homework 3 We mixed 300 ml of a solution of sulfuric acid with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of a solution of potassium hydroxide with a mass fraction of 20% (density of 1.10 g/ml). What volume of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%? Answer: V = 262.9 liters
Homework 4 In 120 ml of a solution of nitric acid with a mass fraction of 7% (density 1.03 g / ml), 12.8 g of calcium carbide was added. What volume of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it? Answer: V = 43.1 ml
Homework 5 When interacting in a sulfuric acid medium, 8.7 g of manganese dioxide with 22.4 g of potassium bromide, bromine was released, the practical yield of which was 88%. What volume (n.c.) of ethylene can react with the resulting amount of bromine? Answer: V = 1.86 L
- Determine the mass of Mg 3 N 2, completely decomposed by water, if 150 ml of a 4% hydrochloric acid solution with a density of 1.02 g / ml was required for salt formation with hydrolysis products.
Show 1) Mg 3 N 2 + 6H 2 O → 3Mg (OH) 2 + 2NH 3
2) Mg(OH) 2 + 2HCl → MgCl 2 + 2H 2 O
3) NH 3 + HCl → NH 4 Cl
n (HCl) \u003d 150 * 1.02 * 0.04 / 36.5 \u003d 0.168 mol
Let x mol of Mg 3 N 2 react. According to equation 1, 3x mol of Mg(OH) 2 and 2x mol of NH 3 were formed. Neutralization of 3x mol of Mg(OH) 2 required 6x mol of HCl (according to equation 2), and neutralization of 2x mol of NH 3 required 2x mol of HCl (according to equation 3), for a total of 8x mol of HCl.
8 x \u003d 0.168 mol,
X \u003d 0.021 mol,
n (Mg 3 N 2) \u003d 0.021 mol,
m (Mg 3 N 2) \u003d M * n \u003d 100 * 0.021 \u003d 2.1 g.
Answer: 2.1 g - Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 8.4% sodium bicarbonate solution. What volume of a 15.6% solution of barium chloride (density 1.11 g / ml) will react with the resulting sodium carbonate? Water evaporation can be neglected.
Show 1) 2NaHCO 3 - t → Na 2 CO 3 + H 2 O + CO 2
2) Na 2 CO 3 + BaCl 2 → BaCO 3 + 2 NaCl
n (NaHCO 3) \u003d 150 * 0.084 / 84 \u003d 0.15 mol
From equation (1) n (NaHCO 3) : n (Na 2 CO 3) = 2:1 => n (Na 2 CO 3) = 0.075 mol.
m (Na 2 CO 3) \u003d 0.075 ∙ 106 \u003d 7.95 g
n (CO 2) \u003d 0.075 mol, m (CO 2) \u003d 0.075 ∙ 44 \u003d 3.3 g
m (solution) \u003d 150 - 3.3 \u003d 146.7 g
ω (Na 2 CO 3) \u003d m (Na 2 CO 3) / m (solution) \u003d 7.95 / 146.7 \u003d 0.0542 or 5.42%
From equation (2) n (Na 2 CO 3) : n (BaCl 2) = 1: 1 => n (BaCl 2) = 0.075 mol.
m (BaCl 2) \u003d n * M \u003d 0.075 * 208 \u003d 15.6 g.
m (solution) \u003d m (BaCl 2) / ω \u003d 15.6 / 0.156 \u003d 100 g
V (solution) \u003d m (solution) / ρ \u003d 100 / 1.11 \u003d 90.1 ml.
Answer: 5.42%, 90.1 ml. - In what mass ratios should 10% sodium hydroxide and sulfuric acid solutions be mixed to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in this solution?
Show 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
Let the mass of NaOH solution be 100 g, m(NaOH)= 10 g,
n(NaOH)=0.25 mol,
n(H 2 SO 4) = 0.125 mol,
m (H 2 SO 4) \u003d 12.25 g,
m (H 2 SO 4 solution) = 122.5 g
Ratio m (NaOH solution): m (H 2 SO 4 solution) = 1:1.2
n(Na 2 SO 4) = 0.125 mol,
m (Na 2 SO 4) \u003d 17.75 g,
m (solution) \u003d 100 + 122.5 g \u003d 222.5 g,
w(Na 2 SO 4)=7.98% - How many liters of chlorine (N.O.) will be released if 26.1 g of manganese (IV) oxide are added to 200 ml of 35% hydrochloric acid (density 1.17 g / ml) when heated? How many grams of sodium hydroxide in a cold solution will react with this amount of chlorine?
Show 1) MnO 2 + 4 HCl → MnCl 2 + 2 H 2 O + Cl 2
2) 2NaOH + Cl 2 → NaCl + NaClO + H 2 O
n (HCl) \u003d 200 * 1.17 * 0.35 / 36.5 \u003d 2.24 mol - in excess
n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 mol - in short supply
According to equation (1) n (Cl 2) \u003d 0.3 mol
V (Cl 2) \u003d 6.72 l
According to equation (2) n(NaOH)= 0.6 mol, m(NaOH)=24 g
Answer: 6.72 l, 24 g - In what volume of water should 11.2 liters of sulfur oxide (IV) (n.a.) be dissolved in order to obtain a solution of sulfurous acid with a mass fraction of 1%? What color will litmus acquire when it is added to the resulting solution?
- What mass of lithium hydride must be dissolved in 100 ml of water to obtain a solution with a mass fraction of hydroxide of 5%? What color will litmus acquire when it is added to the resulting solution?
Show LiH + H 2 O → LiOH + H 2
Let m(LiH)= x g, then m (LiOH)=x*24/8 = 3x g.
m (solution) \u003d m (H 2 O) + m (LiH) - m (H 2)
m (r-ra) \u003d x + 100 - x / 4 \u003d 0.75x + 100
w \u003d m (in-va) * 100% / m (r-ra)
3x/(0.75x+100) = 0.05
3x=0.038x+5
2.96x = 5
x=1.7 g - In what mass of a solution with a mass fraction of Na 2 SO 4 10% should 200 g of Na 2 SO 4 × 10H 2 O be dissolved in order to obtain a solution with a mass fraction of sodium sulfate 16%? What medium will the resulting solution have?
Show Let the mass of the solution be x g. It contains 0.1 x g of Na 2 SO 4.
200 g of crystalline hydrate was added, in which the mass of sodium sulfate is 200*142/322=88.2 g.
(0.1x +88.2) / (x + 200) \u003d 0.16
0.1x +88.2 = 0.16x + 32
0.06x = 56.2
x = 937
Answer: 937 g, neutral. - Gaseous ammonia released by boiling 160 g of a 7% potassium hydroxide solution with 9.0 g of ammonium chloride was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
- Ammonia released by boiling 80 g of a 14% potassium hydroxide solution with 8.03 g of ammonium chloride was dissolved in water. Calculate how many milliliters of 5% nitric acid with a density of 1.02 g / ml will be used to neutralize the resulting ammonia solution.
Show KOH + NH 4 Cl → NH 3 + H 2 O + KCl
NH 3 + HNO 3 → NH 4 NO 3
n (NH 4 Cl) \u003d 8.03 / 53.5 \u003d 0.15 mol
m (KOH) \u003d 80 * 0.14 \u003d 11.2 g
n (KOH) \u003d 11.2 / 56 \u003d 0.2 mol
KOH is in excess.
Further calculations are carried out according to the deficiency
n (NH 4 Cl) \u003d n (NH 3) \u003d 0.15 mol
n (HNO 3) \u003d 0.15 mol
m (HNO 3) \u003d n * M \u003d 0.15 * 63 \u003d 9.45 g
m (p-ra HNO 3) \u003d m (in-va) * 100% / w \u003d 9.45 / 0.05 \u003d 189 g
m = V*ρ
V= m / ρ= 189 / 1.02 = 185.3 ml
Answer: 185.3 ml - Calcium carbide treated with excess water. The released gas occupied a volume of 4.48 liters (N.O.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g / ml will be used to completely neutralize the alkali formed from calcium carbide.
Show 1) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
n (C 2 H 2) \u003d V / V m \u003d 4.48 / 22.4 \u003d 0.2 mol
From equation (1) => n(C 2 H 2) = n(Ca(OH) 2)
n (Ca (OH) 2) \u003d 0.2 mol
2) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O
From equation (2) => n(Ca(OH) 2) : n(HCl) = 1:2 => n(HCl) = 0.4 mol
m(HCl) \u003d n * M \u003d 0.4 * 36.5 \u003d 14.6 g
m (HCl solution) \u003d 14.6 / 0.2 \u003d 73 g
V(HCl solution) = 73/1.1 = 66.4 ml - Calculate what volume of a 10% hydrogen chloride solution with a density of 1.05 g/ml will be used to completely neutralize the calcium hydroxide formed during the hydrolysis of calcium carbide, if the gas released during hydrolysis occupies a volume of 8.96 l (n.o.).
Show CaC 2 + 2 H 2 O → Ca (OH) 2 + C 2 H 2
Ca(OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O
n(C 2 H 2)= V / V n.o.s. = 8.96 / 22.4 = 0.4 mol
n (C 2 H 2) \u003d n Ca (OH) 2 \u003d 0.4 mol
n(HCl) = 0.4 * 2 = 0.8 mol
m (in HCl) \u003d n * M \u003d 0.8 * 36.5 \u003d 29.2 g
w \u003d m (in-va) * 100% / m (r-ra)
m (solution) \u003d m (in-va) * 100% / w \u003d 29.2 / 0.1 \u003d 292 g
m = V*ρ
V= m / ρ = 292 / 1.05 = 278ml
Answer: 278 ml - Aluminum carbide treated with 200 g of a 30% sulfuric acid solution. The methane released at the same time occupied a volume of 4.48 l (n.o.). Calculate the mass fraction of sulfuric acid in the resulting solution.
Show Al 4 C 3 + 6H 2 SO 4 → 2Al 2 (SO 4) 3 + 3CH 4
n(CH 4) \u003d V / Vm \u003d 4.48 / 22.4 \u003d 0.2 mol
m (CH 4) \u003d m * M \u003d 0.2 * 16 \u003d 3.2 g
m(Al 4 C 3) \u003d 1/3 * 0.2 * 144 \u003d 9.6 g
Reacted, according to the equation n (H 2 SO 4) \u003d 0.4 mol,
m (H 2 SO 4) \u003d 0.4 * 98 \u003d 39.2 g.
Initially, m (H 2 SO 4) = m (p-ra) * ω = 200 g * 0.3 = 60 g was added. m (H 2 SO 4) = 60 - 39.2 = 20.8 g remained.
m (H 2 SO 4) \u003d 0.21 * 98 \u003d 20.8 g
m (p-pa) \u003d m (Al 4 C 3) + m (p-pa H 2 SO 4) - m (CH 4)
m(p-pa) = 9.6 g + 200 g - 3.2 g = 206.4 g
ω (H 2 SO 4) \u003d m (H 2 SO 4) / m (solution) \u003d 20.8 / 206.4 * 100% \u003d 10% - When processing aluminum carbide with a solution of hydrochloric acid, the mass of which is 320 g and the mass fraction of HCl 22%, 6.72 l (n.o.) of methane were released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
Show Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 n (CH 4) \u003d V / Vm \u003d b.72 / 22.4 \u003d 0.3 mol;
According to the equation n (HCl) \u003d 4 n (CH 4) \u003d 1.2 mol
m (HCl) \u003d m (solution) * ω \u003d 320 0.22 \u003d 70.4 g;
Reacted m (HCl) \u003d 1.2 36.5 \u003d 43.8 g.
Remaining m(HCl) = 70.4 - 43.8 = 26.6 g.
m (p-pa) \u003d 320 g + m (Al 4 C 3) - m (CH 4),
According to the equation n (Al 4 C 3) \u003d 1/3 n (CH 4) \u003d 0.1 mol;
m (Al 4 C 3) \u003d 0.1 144 \u003d 14.4 g,
m(CH 4) \u003d 0.3 16 \u003d 4.8 g,
m(p-pa) \u003d 320 g + 14.4 g - 4.8 g \u003d 329.6 g.
ω(HCl) = 26.6 / 329.6 100% = 8.07% - Calcium hydride was added to an excess of hydrochloric acid solution (mass of acid solution 150 g, mass fraction of HCl 20%). In this case, 6.72 l (N.O.) of hydrogen were released. Calculate the mass fraction of calcium chloride in the resulting solution.
Show n (CaCl 2) \u003d ½ n (H 2) \u003d 0.15 mol
m (CaCl 2) \u003d 111 * 0.15 \u003d 16.65 g
W (CaCl 2) \u003d m in-va / m solution \u003d 16.65 / 155.7 \u003d 0.1069 or 10.69%
Answer: W (CaCl 2) \u003d 10.69% - 125 ml of 5% lithium hydroxide solution (r = 1.05 g/ml) and 100 ml of 5% nitric acid solution (ρ = 1.03 g/ml) were mixed. Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
Show LiOH + HNO₃ = LiNO₃ + H₂O
m (LiOH solution)= V × ρ = 125 ml × 1.05 g/ml = 131.25 g
m(LiOH) = 131.25 g × 0.05 = 6.563 g
n(LiOH) \u003d m / M \u003d 6.563 / 24 \u003d 0.273 mol
m (solution HNO₃:) = V × ρ = 100 ml × 1.03 g/ml = 103 g
m(HNO₃) = 103 g × 0.05 = 5.15 g
n(HNO₃) = 5.15 / 63 = 0.0817 mol
LiOH is given in excess, the calculation is based on acid.
n(LiNO₃) = 0.0817 mol
m(LiNO₃) = n × M = 0.0817 × 69 = 5.64 g
m (solution obtained) \u003d m (LiOH solution) + m (HNO₃ solution) \u003d 131.25 g + 103 g \u003d 234.25 g
ω(LiNO₃) = 5.64 / 234.25 × 100% = 2.4%
Answer: alkaline, 2.4%; - Phosphorus (V) oxide weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid, and the resulting solution was boiled. What salt and in what quantity is formed if 3.92 g of potassium hydroxide is added to the resulting solution?
- Sulfur oxide (VI) weighing 8 g was dissolved in 110 g of 8% sulfuric acid. What salt and in what quantity is formed if 10.6 g of potassium hydroxide is added to the resulting solution?
Show SO 3 + H 2 O \u003d H 2 SO 4
n \u003d m / MM (SO 3) \u003d 80 g / mol,
n(SO 3) \u003d 8/80 \u003d 0.1 mol.
According to equation (1) n (H 2 SO 4) \u003d n (SO 3) \u003d 0.1 mol,
n(KOH)=10.6/56=0.19 mol.
In the original solution n (H 2 SO 4) \u003d 110 * 0.08 / 98 \u003d 0.09 mol.
After adding sulfur oxide n (H 2 SO 4) \u003d 0.09 + 0.1 \u003d 0.19 mol.
The ratio of alkali and acid is 1:1, which means that an acid salt is formed.
H 2 SO 4 + KOH \u003d KHSO 4 + H 2 O
n (H 2 SO 4) \u003d n (KOH) \u003d n (KHSO 4) \u003d 0.19 mol
Answer: KHSO 4, 0.19 mol. - Ammonia released during the interaction of 107 g of 20% ammonium chloride solution with 150 g of 18% sodium hydroxide solution completely reacted with 60% phosphoric acid to form ammonium dihydrogen phosphate. Determine the mass fraction of sodium chloride in the solution and the required mass of a 60% phosphoric acid solution.
Show NH 4 Cl + NaOH = NaCl + NH 3 + H 2 O
m(NH 4 Cl) \u003d 107 g ∙ 0.2 \u003d 21.4 g
n(NH 4 Cl) \u003d 21.4 g / 53.5 g / mol \u003d 0.4 mol
m(NaOH) = 150 g ∙ 0.18 = 27 g
n(NaOH) = 27 g / 40 g / mol = 0.675 mol, therefore, NaOH is in excess
n(NaCl) = n(NH4Cl) = 0.4 mol
m(NaCl) = 0.4 ∙ 58.5 = 23.4 g
n (NH 3) \u003d n (NH 4 Cl) \u003d 0.4 mol
m (NH 3) \u003d 0.4 ∙ 17 \u003d 6.8 g
m (solution) \u003d m (solution NH 4 Cl) + m (solution NaOH) - m (NH 3) \u003d 107 + 150 - 6.8 \u003d 250.2 g
w(NaCl) = 23.4 / 250.2 = 0.094 or 9.4%
NH 3 + H 3 PO 4 \u003d NH 4 H 2 PO 4
n (NH 3) \u003d n (H 3 PO 4) \u003d 0.4 mol
m (H 3 PO 4) \u003d 98 ∙ 0.4 \u003d 39.2 g
m (solution H 3 PO 4) \u003d 39.2 / 0.6 \u003d 65.3 g - Hydrogen sulfide, released during the interaction of an excess of concentrated sulfuric acid with 1.44 g of magnesium, was passed through 160 g of a 1.5% bromine solution. Determine the mass of the precipitate formed in this case and the mass fraction of acid in the resulting solution.
Show 4Mg + 5H 2 SO 4 \u003d 4MgSO 4 + H 2 S + 4H 2 O
H 2 S + Br 2 \u003d 2HBr + S ↓
n(Mg) \u003d m / M \u003d 1.44 g: 24 g / mol \u003d 0.06 mol
n(H 2 S) = ¼ n(Mg) = 0.015 mol
m(H 2 S) \u003d n * M \u003d 0.015 mol * 34 g / mol \u003d 0.51 mol
m (in-va Br 2) \u003d 160 g * 0.015 \u003d 2.4 g
n(Br 2) \u003d m / M \u003d 2.4 g: 160 g / mol \u003d 0.015 mol
n (HBr) \u003d 2n (Br 2) \u003d 0.03 mol
m(HBr) \u003d n * M \u003d 0.03 mol * 81 g / mol \u003d 2.43 g
n(S) \u003d n (Br 2) \u003d 0.015 mol
m(S) = n * M = 0.015 mol * 32 g/mol = 0.48 g
m (solution) \u003d m (H 2 S) + m (solution Br 2) -m (S) \u003d 0.51 g + 160 g - 0.48 \u003d 160.03 g
W (HBr) \u003d m (HBr) / m (solution) \u003d 2.43 g / 160.03 g \u003d 0.015 or 1.5%
Answer: m (S) \u003d 0.48 g, w (HBr) \u003d 1.5% - Chlorine reacted without residue with 228.58 ml of 5% NaOH solution (density 1.05 g/ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
Show 6NaOH + 3Cl 2 = 5NaCl + NaClO 3 + 3H 2 O (at t)
m solution \u003d 228.58 ∙ 1.05 \u003d 240 g;
m(NaOH) \u003d 240 ∙ 0.05 \u003d 12g.
n(NaOH) = 12/40 = 0.3 mol;
n(Cl 2) = 0.15 mol;
n(NaCl) = 0.25 mol;
n(NaClO 3) = 0.05 mol
m (NaCl) \u003d 58.5 ∙ 0.25 \u003d 14.625 g;
m (NaClO 3) \u003d 106.5 ∙ 0.05 \u003d 5.325 g:
m solution \u003d 240 + m (Cl 2) \u003d 240 + 71 ∙ 0.15 \u003d 240 + 10.65 \u003d 250.65 g
W(NaCl) = 14.625 / 250.65 = 0.0583 or 5.83%
W (NaClO 3) \u003d 5.325 / 250.65 \u003d 0.0212 or 2.12% - Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ = 1.153 g/ml). For complete binding of the products, 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkali in the used solution.
Show 3Cu + 8HNO 3 \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O
m (HNO 3) \u003d 100 ∙ 0.3 ∙ 1.153 \u003d 34.59g
n(HNO 3) = 34.59/ 63 = 0.55 mol, n(Cu) = 6.4/ 64 = 0.1 mol
n (HNO 3) g = 0.55 - 8/3 ∙ 0.1 = 0.28 mol
Cu (NO 3) 2 + 2 NaOH \u003d Cu (OH) 2 + 2 NaNO 3
HNO 3 + NaOH \u003d NaNO 3 + H 2 O
n(NaOH) \u003d n (HNO 3) g + 2n (Cu (NO 3) 2) \u003d 0.28 + 0.1 ∙ 2 \u003d 0.48 mol
m (NaOH) \u003d 0.48 ∙ 40 \u003d 19.2 g.
W(NaOH) = 19.2/ 200 = 0.096 or 9.6% - In 60 g of 18% orthophosphoric acid, 2.84 g of phosphorus (V) oxide was dissolved and the resulting solution was boiled. What salt and in what quantity is formed if 30 g of sodium hydroxide is added to the resulting solution?
Show 1) 3H 2 O + P 2 O 5 → 2H 3 PO 4
In the original solution m (H 3 PO 4) \u003d m (solution) * ω \u003d 60 * 0.18 \u003d 10.8 g.
n(P 2 O 5) = m/M = 2.84/142 = 0.02 mol
As a result of the reaction, m (H 3 PO 4) \u003d 0.04 * 98 \u003d 3.92 g was formed
Total m (H 3 PO 4) \u003d 3.92 + 10.8 \u003d 14.72 g.
n (H 3 PO 4) \u003d m / M \u003d 14.72 / 98 \u003d 0.15 mol
n(NaOH) \u003d m / M \u003d 30/40 \u003d 0.75 mol - in excess, salt is average.
2) 3NaOH + H 3 PO 4 → Na 3 PO 4 + 3H 2 O
According to equation (2) n (Na 3 PO 4) \u003d n (H 3 PO 4) \u003d 0.15 mol
m (Na 3 PO 4) \u003d 0.15 * 164 \u003d 24.6g
Answer: 24.6 g - Ammonia with a volume of 4.48 l (n.o.) was passed through 200 g of a 4.9% solution of phosphoric acid. Name the salt formed as a result of the reaction, and determine its mass.
- 5.6 l (n.o.) of hydrogen sulfide reacted without residue with 59.02 ml of 20% KOH solution (density 1.186 g/ml). Determine the mass of the salt produced by this chemical reaction.
Show m (p-ra KOH) \u003d V * ρ \u003d 1.186 * 59.02 \u003d 70 g,
m (KOH) \u003d m (p-raKOH) * ω \u003d 70 g * 0.2 \u003d 14 g,
n(KOH) \u003d m / M \u003d 14/56 \u003d 0.25 mol,
n(H 2 S) \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.
The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acid salt is formed - hydrosulfide according to the reaction: H 2 S + KOH \u003d KHS + H 2 O
According to the equation n(KHS) = 0.25 mol,
m(KHS) \u003d M * n \u003d 72 0.25 \u003d 18 g.
Answer: 18 - To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution (density 1.20 g/ml) was used. Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
Show COOH + KOH \u003d NSOOK + H 2 O
CH 3 COOH + KOH \u003d CH 3 COOK + H 2 O
m (p-paKOH) \u003d V (p-pa) * ρ \u003d 35 * 1.2 \u003d 42 g.
m(KOH) \u003d m (p-pa) * ω (KOH) \u003d 42 * 0.2 \u003d 8.4 g
n(KOH) \u003d m (KOH) / M (KOH) \u003d 8.4 / 56 \u003d 0.15 mol
Let n (HCOOH) \u003d x mol, and n (CH 3 COOH) \u003d y mol.
m (HCOOH) \u003d n (HCOOH) * M (HCOOH) \u003d x * 46 g
m (CH 3 COOH) \u003d n (CH 3 COOH) * M (CH 3 COOH) \u003d y * 60 g
Let's make a system of equations:
x + y = 0.15
60y + 46x = 7.6
We solve the system: x \u003d 0.1 mol, y \u003d 0.05 mol
m (CH 3 COOH) \u003d n (CH 3 COOH) * M (CH 3 COOH) \u003d 0.05 * 60 \u003d 3 g.
ω(CH 3 COOH) = m(CH 3 COOH)/m(mixture) = 3/7.6 = 0.395 or 39.5%.
Answer: 39.5% - 100 ml of 30% perchloric acid solution (r = 1.11 g/ml) and 300 ml of 20% sodium hydroxide solution (r = 1.10 g/ml) were mixed. How many milliliters of water should be added to the resulting mixture so that the mass fraction of sodium perchlorate in it would be 8%?
Show HClO 4 + NaOH = NaClO 4 + H 2 O
m (p-p NaOH) \u003d V * ρ \u003d 300 * 1.10 \u003d 330 g.
n (NaOH) \u003d m (p-p NaOH) * ω / M \u003d 330 * 0.2 / 40 \u003d 1.65 mol - in excess.
m (solution HClO 4) \u003d V * ρ \u003d 100 * 1.11 \u003d 111 g.
n (HClO 4) \u003d 111 * 0.3 / 100.5 \u003d 0.331 mol,
According to the equation n (HClO 4) \u003d n (NaClO 4) \u003d 0.331 mol,
m (NaClO 4) \u003d n * M \u003d 0.331 * 122.5 \u003d 40.5 g.
Let the mass of added water be x g.
40.5 / (111 + 330 + x) \u003d 0.08
whence x = 65.3 g.
V(H 2 O)=65.3 ml.
Answer: 65.3 ml - To 100 ml of a 5% hydrochloric acid solution (density 1.02 g/ml) was added 6.4 g of calcium carbide. How many milliliters of 15% nitric acid (density 1.08 g/ml) should be added to the resulting mixture to completely neutralize it?
Show 1) CaC 2 + 2HCl \u003d CaCl 2 + C 2 H 2
2) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
3) Ca(OH) 2 + 2HNO 3 = Ca(NO 3) 2 + 2H 2 O
n(HCl) \u003d m (HCl) / M (HCl) \u003d Vp-ra (HCl) * (HCl) * p (HCl) / M (HCl) \u003d 100 * 0.05 * 1.02 / 36.5 \u003d 0.14 mol,
n (CaC 2) \u003d m / M \u003d 6.4 / 64 \u003d 0.1 mol.
According to equation (1) n (CaC 2) : n (HCl) = 1: 2 => CaC 2 - in excess.
Reacted n (CaC 2) = n (HCl) / 2 = 0.07 mol.
It remains n (CaC 2) \u003d 0.1 - 0.07 \u003d 0.03 mol.
According to equation (2) n (CaC 2) \u003d n (Ca (OH) 2) \u003d 0.03 mol.
By equation (3) n(Ca(OH)2) : n(HNO3) = 1: 2 =>
n (HNO 3) \u003d 2n (Ca (OH) 2) \u003d 0.06 mol.
Vr-ra (HNO 3) \u003d m (r-ra) / ρ
m (solution) \u003d m (HNO 3) / ω \u003d 0.06 * 63 / 0.15 \u003d 25.2 g,
V (p-raHNO 3) \u003d 25.2 / 1.08 \u003d 23.3 ml.
Answer: 23.3 ml - Sodium nitrite weighing 13.8 g was introduced by heating into 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium chloride in the resulting solution?
Show NaNO 2 + NH 4 Cl \u003d N 2 + NaCl + 2H 2 O
n (NaNO 2) \u003d 13.8 / 69 \u003d 0.2 mol
n (NH 4 Cl) \u003d 220 0.1 / 53.5 \u003d 0.41 mol
NH 4 Cl - in excess n (N 2) \u003d n (NaNO 2) \u003d 0.2 mol
V (N 2) \u003d 0.2 mol 22.4 l / mol \u003d 4.48 l
Calculate the mass of ammonium chloride remaining in excess:
n(NH 4 Cl) g = 0.41 - 0.2 = 0.21 mol
m (NH 4 Cl) g = 0.21 53.5 = 11.2 g Calculate the mass fraction of ammonium chloride:
m(p-pa) \u003d 13.8 + 220 - 0.2 28 \u003d 228.2 g
ω (NH 4 Cl) \u003d 11.2 / 228.2 \u003d 0.049 or 4.9% Answer:
V (N 2) \u003d 4.48 l
ω(NH 4 Cl) \u003d 4.9% - Potassium nitrite weighing 8.5 g was introduced by heating into 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (N.O.) of nitrogen will be released in this case and what is the mass fraction of ammonium bromide in the resulting solution?
Show KNO 2 + NH 4 Br \u003d N 2 + KBr + 2H 2 O
m (in-va NH 4 Br) \u003d 270 g * 0.12 \u003d 32.4 g
n(NH 4 Br) \u003d m / M \u003d 32.4 g: 98 g / mol \u003d 0.33 mol
n(KNO 2) \u003d m / M \u003d 8.5 g: 85 g / mol \u003d 0.1 mol
n(NH 4), proreact. with KNO 2 \u003d 0.33 mol - 0.1 mol \u003d 0.23 mol (since n (KNO 2) : n (NH 4 Br) \u003d 1: 1)
m (NH 4 Br) remaining in the final solution \u003d n * M \u003d 0.23 mol * 98 g / mol \u003d 22.54 g
n (N 2) \u003d n (KNO 2) \u003d 0.1 mol
(N 2) \u003d n * M \u003d 0.1 mol * 28 g / mol \u003d 2.8 g
V (N 2) \u003d n * M \u003d 0.1 mol * 22.4 l / mol \u003d 2.24 l
m (final solution) \u003d m (KNO 2) + m (NH 4 Br solution) - m (N 2) \u003d 8.5 g + 270 g - 2.8 g \u003d 275.7 g
W(NH 4 Br in final solution) = 22.54 g: 275.7 g = 8%
Answer: V (N 2) \u003d 2.24 l; W(NH 4 Br) = 8% - Mixed 300 ml of a solution of sulfuric acid with a mass fraction of 10% (density 1.05 g/ml) and 200 ml of a solution of potassium hydroxide with a mass fraction of 20% (density of 1.10 g/ml). How many milliliters of water should be added to the resulting mixture so that the mass fraction of salt in it is 7%?
- In 120 ml of a solution of nitric acid with a mass fraction of 7% (density 1.03 g/ml) made 12.8 g of calcium carbide. How many milliliters of 20% hydrochloric acid (density 1.10 g/ml) should be added to the resulting mixture to completely neutralize it?
Show 1) CaC 2 + 2HNO 3 \u003d Ca (NO 3) 2 + C 2 H 2.
n (CaC 2) \u003d 12.8 / 64 \u003d 0.2 mol
n (HNO 3) \u003d (0.07 1.03 120) / 63 \u003d 0.137 mol
CaC 2 - excess. 2) CaC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2.
n (Ca (OH) 2) \u003d 0.2 - 0.137 / 2 \u003d 0.13 mol 3) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O.
n(HCl) \u003d 0.13 2 \u003d 0.26 mol
m (solution) \u003d m (HCl) / w \u003d (0.26 36.5) / 0.2 \u003d 47.45 g
V (solution HCl) \u003d m (solution) / ρ \u003d 47.45 / 1.10 \u003d 43.1 ml.
Answer: 43.1 ml. - To the solution obtained by adding 4 g of potassium hydride to 100 ml of water was added 100 ml of a 39% solution of nitric acid (r = 1.24 g/ml). Determine the mass fractions of all substances (including water) in the final solution.
Show KH + H 2 O \u003d KOH + H 2
n(KH) = 4 g: 40 g/mol = 0.1 mol
n(H 2 O) \u003d 100 g: 18 g / mol \u003d 5.6 mol
KH is in short supply, then n (KOH) \u003d n (KH) \u003d 0.1 molKOH + H 2 NO 3 \u003d KNO 3 + H 2 O
a) m(KOH) = 0.1 * 56 = 5.6
b) m in-va (HNO 2) \u003d 100 ml * 1.24 g / ml * 0.39 \u003d 48.36 g
n (HNO 3) \u003d 48.36 g: 63 g / mol \u003d 0.77 mol
HNO 3 in excess, n excess (HNO 3) \u003d 0.77 - 0.1 \u003d 0.67 mol
m (HNO 3) \u003d 0.67 * 63 \u003d 42.21 g
m (solution) \u003d 4g + 100g + 124g - 0.2g \u003d 227.8g
3) W (KNO 3) \u003d m (KNO 3) : m (p-pa) \u003d (0.1 mol * 101 g / mol) 227.8 g * 100% \u003d 4.4%
W (HNO 3) \u003d 42.21: 227.8 * 100% \u003d 18.5
W (H 2 O) \u003d 100% - (W (KNO 3) + W (HNO 3)) \u003d 77.1%
Answer:
W(KNO 3) = 4.4%
W(HNO 3) \u003d 18.5%
W(H 2 O) = 77.1%1) 2Na 2 O 2 + 2H 2 O \u003d 4NaOH + O 2
2) 2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
m (solution H 2 SO 4) \u003d 300 * 1.08 \u003d 324 g
m (H 2 SO 4) \u003d 0.1 * 324 \u003d 32.4 g
n (H 2 SO 4) \u003d 32.4 / 98 \u003d 0.33 mol
n(NaOH): n(H 2 SO 4) = 2:1 => n(NaOH) = 0.33 * 2 = 0.66 mol
n(Na 2 O 2): n(NaOH) = 1:2 => n(Na 2 O 2) = 0.66/2 = 0.33 mol
m (Na 2 O 2) \u003d n * M \u003d 0.33 * 78 \u003d 25.7 g
n(Na 2 O 2): n(O 2) = 2:1 => n(O 2) = 0.33/2 = 0.165 mol
V (O 2) \u003d 0.165 * 22.4 \u003d 3.7 l
Answer: m (Na 2 O 2) \u003d 25.7 g; V (O 2) \u003d 3.7 l - When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the initial solution, by heating which you can get an 8% solution of potassium carbonate.
- When interacting in a sulfuric acid medium, 17.4 g of manganese dioxide with 58 g of potassium bromide, bromine was released at a 77% yield. What volume (N.O.) of propene can react with the resulting amount of bromine?
- Carbon dioxide with a volume of 5.6 l (n.o.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ = 1.22 g/ml). Determine the composition and mass fractions of substances in the resulting solution.
- Aluminum carbide was dissolved in a 15% sulfuric acid solution weighing 300 g. The methane released at the same time occupied a volume of 2.24 l (n.o.). Calculate the mass fraction of sulfuric acid in the resulting solution.
- In an excess of oxygen, 8 g of sulfur were burned. The resulting gas was passed through 200 g of 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
- A mixture of aluminum and iron filings was treated with an excess of dilute hydrochloric acid, and 8.96 L (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of iron in the initial mixture.
- A mixture of magnesium and zinc filings was treated with an excess of dilute sulfuric acid, and 22.4 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of sodium hydroxide solution, then 13.44 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
2KHCO 3 + H 2 SO 4 = K 2 SO 4 + 2CO 2 + 2H 2 O (2) Na 2 CO 3 + 2HCl \u003d NaCl + CO 2 + H 2 O
NaHCO 3 + HCl \u003d NaCl + CO 2 + H 2 O
NaHCO 3 + NaOH \u003d Na 2 CO 3 + H 2 O
1) findamountNaHCO3
n(NaOH)=(NaHCO 3)=m(NaOH) . ω (NaOH) / M (NaOH) \u003d 80 * 0.1 / 40 \u003d 0.2 mol
2) findmassNaHCO3
m(NaHCO 3)=n(NaHCO 3)*M(NaHCO 3)=0.2. 84=16.8 g
3) find the amount of Na 2 CO 3
n(HCl) = m(HCl) . ω (HCl) / M (HCl) \u003d 73 * 0.2 / 36.5 \u003d 0.4 mol
n (Na 2 CO 3) \u003d (n (HCl) - n (NaOH)) / 2 \u003d 0.1 mol
4) find the massNa 2 CO 3
m(Na 2 CO 3)=n. M = 0.1. 106 = 10.6 g
5) find the mass fractionNa2CO3
ω (Na 2 CO 3) \u003d m (Na 2 CO 3) / m (mixture) . 100%=38.7%
m(mixture).=m(Na 2 CO 3)+m(NaHCO 3)=27.4 g
Answer: ω (Na 2 CO 3) \u003d 38.7%1) 2Na + 2H 2 O → 2NaOH + H 2
2) Na 2 O + H 2 O → 2NaOH
1) find the quantityNa
n(H 2) \u003d V / Vm \u003d 4.48 / 22.4 \u003d 0.2 mol
According to equation (1) n (Na) \u003d n 1 (NaOH) \u003d 2. n(H 2) = 0.4 mol
2) find the massNa
m(Na) = n . M = 0.4. 23 = 9.2 g
3) find the quantityNaOH
n(NaOH) = m(p-p) . ω/M(NaOH) = 240 . 0.1/40 = 0.6 mol
4) find the quantityNa 2 O
n (Na 2 O) \u003d (n (NaOH) - n 1 (NaOH)) / 2 \u003d 0.1 mol
m(Na 2 O)= n. M = 0.1. 62 = 6.2g
5) findmmass shareNa
ω (Na) = m(Na)/m(mixtures) . 100%=59.7%
m(mixture)=m(Na)+m(Na 2 O)=15.4
Answer: ω (Na)=59,7%
1) Find the totalH2SO4
n 1 (H 2 SO 4) \u003d m. ω/M=490 . 0.4/98=2 mol
M(H 2 SO 4) \u003d 2 + 32 + 64 \u003d 98 g / mol
2) Let's findH2SO4reactedwithNa2CO3
n (Na 2 CO 3) \u003d n (Na 2 CO 3. 10H 2 O) \u003d m / M \u003d 143/286 \u003d 0.5 mol
M(Na 2 CO 3. 10H 2 O) \u003d (46 + 12 + 48) + (10. 18) \u003d 286 g / mol
According to the equation n (H 2 SO 4) \u003d n (Na 2 CO 3) \u003d 0.5 mol
3) findH 2 SO 4 reacted withNaOH
n (H 2 SO 4) \u003d n 1 -n \u003d 2-0.5 \u003d 1.5 mol
4) find the mass of NaOH
n(NaOH)=2n(H 2 SO 4)=2. 1.5=3 mol
m(NaOH)=n. M=3 . 40=120 g
M(NaOH)=23+16+1=18 g/mol
5) find the mass fractionNaOH
ω (NaOH) \u003d m (substance) / . m(p-p)*100%=120/1200 . 100%=10%
Answer: m(NaOH)=120g; ω(NaOH)=10%
48) A mixture of magnesium and aluminum filings was treated with an excess of dilute hydrochloric acid, and 11.2 liters (n.o.) of hydrogen were released. If the same mass of the mixture is treated with an excess of potassium hydroxide solution, then 6.72 liters (N.O.) of hydrogen will be released. Calculate the mass fraction of magnesium in the initial mixture.
49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of hydrobromic acid (ρ = 1.12 g / ml) with a mass fraction of 20%. What is the mass fraction of hydrogen bromide in the resulting solution?
In this problem, there are two parallel reactions involving a mixture of substances.
Problem 3.9.
To neutralize 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% potassium hydroxide solution was used (p = 1.2 g/ml). Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
Given:
mass of acid mixture: m (acid mixture) = 7.6 g;
volume of KOH solution: V solution (KOH) = 35 ml;
mass fraction of KOH in the original solution: (KOH) in ref. p-re = 20%;
density of the initial KOH solution: p ref. solution (KOH) = 1.2 g / ml.
To find: mass of acetic acid: m(CH 3 COOH);
mass fraction of CH 3 COOH in the initial mixture: (CH 3 COOH) = ?
Decision:
When a mixture of acids interacts with potassium hydroxide, two reactions occur simultaneously:
CH 3 COOH + KOH → CH 3 COOK + H 2 O
HCOOH + KOH → LCOOK + H 2 O
The solution of the problem is possible according to the counter algorithm with the compilation of a mathematical equation. The solution scheme can be represented as follows:
1. Denote the mass of acetic acid in the initial mixture by the value "a":
m (CH 3 COOH) \u003d a g.
Then the mass of formic acid can be determined by the difference:
m (HCOOH) \u003d m (mixtures of acids) - m (CH 3 COOH) \u003d (7.6 - a) g.
2. Using the equation for the reaction of acetic acid with an alkali solution, we determine the mass of KOH that was consumed in this interaction.
3. Similarly, according to the reaction equation with formic acid, we find the mass of KOH, which was consumed in the second interaction.
4. Find total weight KOH, which was consumed in two reactions:
m (KOH) \u003d m (KOH) in the district with ux to the one + m (KOH) in the district with mur to the one \u003d
= 0.933. a + (9.25 - 1.217. a) \u003d (9.25 - 0.284. a) g.
5. Find the mass of KOH, which was contained in 200 ml of the initial solution:
6. Equate the expression obtained in the 4th action to the value of the KOH mass from the 5th action:
9.25 - 0.284. a = 8.4.
Got a math equation. Solving it, we find the value of the value "a":
The value of "a" we denoted the mass of acetic acid:
m (CH 3 COOH) \u003d 3 g.
7. Find the mass fraction of CH 3 COOH in the initial mixture of acids
Answer: m(CH 3 COOH) = 3 g; (CH 3 COOH) = 39.5%.
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