# A sample weighing 40 g, containing CaC2 and AL4C3 in the same chemical amount

A sample weighing 40 g, containing CaC2 and AL4C3 in the same chemical amount, was treated with excess water. How much gas is released in this case?

To solve, we compose the equations of the processes:

CaC2 + 2H2O = Ca (OH) 2 + C2H2 – hydration of calcium carbide, acetylene obtained;
Al4C3 + 12H2O = 3CH4 + 4Al (OH) 3 – methane formed;
Calculations:
M (CaC2) = 64 g / mol;

Y (CaC2) = m / M = 20/64 = 0.3 mol;

V (C2H2) = 0.3 * 22.4 = 6.72 L;

M (Al4C3) = 143.6 g / mol;

Y (Al4C3) = m / M = 20 / 143.6 = 0.14 mol.

Proportion:
0.14 mol (Al4C3) – X mol (CH4);

-1 mol -3 mol hence, X mol (CH4) = 0.14 * 3/1 = 0.42 mol.

Find the volume of the product:
V (CH4) = 0.42 * 22.4 = 9.4 L

Answer: the volume of acetylene was 6.72 liters, methane – 9.4 liters

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