27.2 g of a mixture of calcium and aluminum carbides was treated with acid, and 11.2 liters

27.2 g of a mixture of calcium and aluminum carbides was treated with acid, and 11.2 liters of a mixture of gases were obtained. Determine the volume fraction of acetylene in the mixture.

Let’s write down the reaction equations:
CaC2 + 2HCl = CaCl2 + C2H2.
Al4C3 + 12HCl = 4AlCl3 + 3CH4.
Let’s find the number of moles of a mixture of gases:
n = V / Vm = 11.2 / 22.4 = 0.5.
Let’s denote the unknown amount of gases through X and Y.
n (CaC2) = n (C2H2) = X.
n (Al4C3) = 3n (CH4) = Y.
First equation:
X + 3Y = 0.5.
Second equation x moles of calcium carbide and y moles of aluminum carbide.
64X + 144 * Y = 27.2.
Let’s solve the system of equations:
X = 0.5-3Y.
32 – 192Y + 144Y = 27.2.
-48y = -4.8.
y = 0.1.
X = 0.2.
Now we will find the volume fraction of acetylene in the mixture.
V (C2H2) = 0.2 / 0.5 = 40%.