What volume of methane will be released when 10 g of aluminum carbide (Al4C3) interacts with 10 g of water?

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

Let’s find the amounts of water and aluminum carbide:

n (H2O) = m (H2O) / M (H2O) = 10/18 = 0.556 mol;

n (Al4C3) = m (Al4C3) / M (Al4C3) = 10/144 = 0.0694 mol;

There is a shortage of water, so the calculation is based on it.

n (CH4) = 0.25n (H2O) = 0.556 * 0.25 = 0.139 mol;

V (CH4) = n (CH4) * VM = 0.139 * 22.4 = 3.114 l.

Answer: V (CH4) = 3.114 l.