Calculate the mass of aluminum carbide (AL4C3) required to produce 134.4 liters of methane under normal conditions.

Let’s solve the problem:
1. Al4C3 + 12H2O = 3CH4 + 4Al (OH) 3 – OBP, methane was obtained from aluminum carbide;
2. Let’s calculate the molar masses:
M (Al4C3) = 143.6 g / mol;
M (CH4) = 134.4 g / mol.
3. Let’s make the proportions according to the equation:
1 mol of gas at normal level – 22.4 liters;
X mol (CH4) – 134.4 liters. hence, X mol (CH4) = 1 * 134.4 / 22.4 = 6 mol;
X mol (Al4C3) – 6 mol (CH4);
-1 mol -3 mol from here, X mol (Al4C3) = 1 * 6/3 = 2 mol.
4. Find the mass of Al4C3:
m (Al4C3) = Y * M = 2 * 143.6 = 287.2 g.
Answer: The hydration process requires 287.2 g of aluminum carbide.