24 g of technical calcium carbide was treated with an excess of hydrochloric acid. Determine the volume

24 g of technical calcium carbide was treated with an excess of hydrochloric acid. Determine the volume of the resulting gas if the mass fraction of impurities in technical calcium carbide is 10% and the product yield is 80%

1. The interaction of calcium carbide with hydrochloric acid describes the reaction:

CaC2 + 2 HCl = CaCl2 + C2H2 ↑;

2. Calculate the mass of calcium carbide in the sample:

m (CaC2) = m (sample) – m (impurities);

m (impurities) = w (impurities) * m (sample) = 0.1 * 24 = 2.4 g;

m (CaCl2) = 24 – 2.4 = 21.6 g;

3. Determine the chemical amount of carbide:

n (CaCl2) = m (CaCl2): M (CaCl2);

M (CaCl2) = 40 + 2 * 35.5 = 111 g / mol;

n (CaCl2) = 21.6: 111 = 0.1946 mol;

4. Let us establish the theoretical amount of ethylene obtained:

ntheor (C2H2) = n (CaCl2) = 0.1946 mol;

5. Knowing the yield of the reaction product, we calculate the practical amount of ethylene:

npr (C2H2) = ν * ntheor (C2H2) = 0.8 * 0.1946 = 0.1557 mol;

6. Let’s calculate the volume of ethylene:

V (C2H2) = npr (C2H2) * Vm = 0.1557 * 22.4 = 3.49 liters.

Answer: 3.49 liters.