Calculate the volume of methane that will be released when treating 11.7 g of aluminum carbide with water.

To solve the problem, you need to draw up an equation:
1. Al4C3 + 12H2O = 3CH4 + 4Al (OH) 3 – hydration, methane and aluminum oxide are released;
2. Calculations:
M (Al4C3) = 143.6 g / mol;
M (CH4) = 16 g / mol.
3. Determine the amount of the original substance:
Y (Al4C3) = m / M = 11.7 / 143.6 = 0.08 mol.
4. Proportion:
0.08 mol (Al4C3) – X mol (CH4);
-1 mol -3 mol from here, X mol (CH4) = 0.08 * 3/1 = 0.24 mol.
5. Find the volume of the product:
V (CH4) = 0.24 * 22.4 = 5.376 l.
Answer: Methane produced with a volume of 5.376 liters.