Technical calcium carbide contains calcium sulfide and other impurities that do not react with water.

Technical calcium carbide contains calcium sulfide and other impurities that do not react with water. When processing 1.38 g of a technical preparation with an excess of boiling water, 470.4 cm3 of hydrogen sulfide was collected, after passing it through an excess of CuSO4 solution, 0.096 g of precipitate falls out. Find w (CaC2) in the original preparation.

Given:
m tech. (CaC2) = 1.38 g
V (gas) = 470.4 cm3 = 0.4704 l
m (sediment) = 0.096 g

To find:
ω (CaC2) -?

Decision:
1) CaC2 + 2H2O => Ca (OH) 2 + C2H2 ↑;
CaS + 2H2O => Ca (OH) 2 + H2S ↑;
H2S + CuSO4 => CuS ↓ + H2SO4;
2) n (gas) = V (gas) / Vm = 0.4704 / 22.4 = 0.021 mol;
3) n (CuS) = m (CuS) / M (CuS) = 0.096 / 96 = 0.001 mol;
4) n (H2S) = n (CuS) = 0.001 mol;
5) n (C2H2) = n (gas) – n (H2S) = 0.021 – 0.001 = 0.02 mol;
6) n (CaC2) = n (C2H2) = 0.02 mol;
7) m clean. (CaC2) = n (CaC2) * M (CaC2) = 0.02 * 64 = 1.28 g;
8) ω (CaC2) = m pure. (CaC2) * 100% / m tech. (CaC2) = 1.28 * 100% / 1.38 = 92.75%.

Answer: The mass fraction of CaC2 is 92.75%.