How many grams of acetaldehyde can be obtained from 75.3 g of calcium carbide containing 15% impurities?

To solve, we write down the equations of the process:

m = 75.3 g; 15% impurity

CaC2 + 2H2O = Ca (OH) 2 + C2H2 – hydration, acetylene obtained;

С2Н2 + Н2О = Н2С = СНОН (vinyl alcohol) = СН3 – СОН – acetaldehyde was formed;
Calculations:
M (CaC2) = 64 g / mol;

M (C2H4O) = 44 g / mol;

m (CaC2) = 75.3 * (1 – 0.15) = 64 g (weight without impurity);

Y (CaC2) = m / M = 64/64 = 1 mol;

Y (C2H4O) = 1 mol so the amount of substances is equal to 1 mol.

Find the mass of the product:
m (C2H4O) = Y * M = 1 * 44 = 44 g

Answer: the mass of acetaldehyde is 44 g