# 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.

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