Find the volume of gas that is formed by the interaction of 50 g of calcium carbide containing 5% impurities and 50 g of water.

1. Let’s write the equation of interaction of calcium carbide with water:

CaC2 + 2H2O = Ca (OH) 2 + C2H2;

2.Calculate the mass of the carbide:

m (CaC2) = (1 – w (CaC2)) * m (solution) = 0.95 * 50 = 47.5 g;

3.Calculate the chemical quantities of the reacting substances:

n (CaC2) = m (CaC2): M (CaC2);

M (CaC2) = 40 + 12 * 2 = 64 g / mol;

n (CaC2) = 47.5: 64 = 0.7422 mol;

n (H2O) = m (H2O): M (H2O) = 50: 18 = 2.7778 mol;

4.determine the amount of acetylene for the substance in short supply:

n (C2H2) = n (CaC2) = 0.7422 mol;

5. find the volume of the formed acetylene:

V (C2H2) = n (C2H2) * Vm = 0.7422 * 22.4 = 16.63 liters.

Answer: 16.63 liters.