Find the mass of aluminum carbide required to obtain 112 liters of methane if the fraction of the reaction

Find the mass of aluminum carbide required to obtain 112 liters of methane if the fraction of the reaction product yield is 80% of the theoretically possible.

Let’s execute the solution:

In accordance with the condition of the problem, we will write the data:
Xr -? V = 112 l; W = 80%

Al4C3 + 12H2O = 3CH4 + 4Al (OH) 3 – hydration of aluminum carbide, methane obtained;

Calculations:
M (Al4C3) = 143.6 g / mol;

M (CH4) = 16 g / mol.

Proportion:
1 mol of gas at normal level – 22.4 liters;

X mol (CH4) -112 L from here, X mol (CH4) = 1 * 112 / 22.4 = 5 mol.

Find the volume of the product:
V (CH4) = 5 * 22.4 = 112 liters (theoretical volume);

V (CH4) = 0.80 * 112 = 89.6 L

Answer: the volume of methane is 89.6 liters.