# What volume of methane will be released: a) during hydrolysis of 72 g of Al4C3; b) from 4.1 g of anhydrous sodium acetate?

Solution:

a) Al4C3 + 12H2O = 4Al (OH) 3 + 3CH4;

Determine the number of moles of Al4C3 if m (AL4C3) = 72 g;

Y (Al4C3) = m (Al4C3) / M (Al4C3); Y (Al4C3) = 4.1 / 82 = 0.05 mol.

Let’s make the proportion:

0.05 mol (Al4C3) – X mol (CH4);

hence, X mol (CH4) = 0.05 * 1/1 = 0.05 mol.

Let’s calculate the volume of methane: V (CH4) = 0.05 * 22.4 = 1.12 L;

Answer: during hydrolysis of aluminum carbide, methane is released in a volume of 1.12 liters.

b) We compose the methane synthesis equation: CH3COONa + NaOH = CH4 + Na2CO3;

Let us determine the number of mol of CH3COONa: Y (CH3COONa) = 4.1 / 82 = 0.05 mol.

Let’s write down the proportion and determine the amount of CH4 mol.

0.05 mol (CH3COONa) – X mol (CH4)

Thus, X mol (CH4) = 0.05 * 1/1 = 0.05.

We calculate the volume of CH4: V (CH4) = 0.05 * 22.4 = 1.12L

Answer: methane with a volume of 1.12 liters was released.