The gas released during the processing of 100 grams of an alloy of copper and zinc with an excess of hydrochloric acid

| April 10, 2021 | education

The gas released during the processing of 100 grams of an alloy of copper and zinc with an excess of hydrochloric acid, when heated, completely reduced the Fe (III) oxide, while the mass of the oxide decreased by 9.6 g. Find the mass fractions of metals in the mixture.

Given:
m (Cu + Zn) = 100 g
m (Fe2O3) = 9.6 g
To find:
ω (Сu) -?
ω (Zn) -?
Decision.
M (Fe2O3) = 160 g / mol
M (Zn) = 65 g / mol
Vm = 22.4 l / mol
Find the volume of hydrogen:
9.6 g-x l
Fe2O3 + 3H2 = 2Fe + 3H2O
1 mole; 3 mol
160 g; 67.2 l
x = V (H2) = 9.6 g * 67.2 l / 160 g = 4.032 l
Find the mass of zinc:
x g-4,032 l
Zn + 2HCl = ZnCl2 + H2 ↑
1 mole; 1 mol-H2
65 g-22.4 L
x = m (Zn) = 4.032 L × 65 g / 22.4 L = 11.7 g
m (Cu) = 100 g – 11.7 g = 88.3 g
ω (Cu) = 88.3 g / 100 g × 100% = 88.3%
ω (Zn) = 11.7 g / 100 g × 100% = 11.7%
Answer: ω (Сu) = 88.3%; ω (Zn) = 11.7%



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