Determine the volume of methane that can be obtained from 36 g of aluminum carbide with a mass fraction of impurities of 20%.

Given:
m tech. (Al4C3) = 36 g
ω approx. = 20%
Find: V (CH4) -?
1) Al4C3 + 12H2O => 4Al (OH) 3 + 3CH4;
2) ω (Al4C3) = 100% – ω approx. = 100% – 20% = 80%;
3) m clean. (Al4C3) = ω * m tech. / 100% = 80% * 36/100% = 28.8 g;
4) n clean. (Al4C3) = m pure. / M = 28.8 / 144 = 0.2 mol;
5) n (CH4) = n pure. (Al4C3) * 3 = 0.2 * 3 = 0.6 mol;
6) V (CH4) = n * Vm = 0.6 * 22.4 = 13.4 liters.
Answer: The volume of CH4 is 13.4 liters.