Al4C3 + 12HCl = 4AlCl3 + 3CH4 ↑
From the condition of the problem m (Al4C3) = 86.4 g. We find the amount of Al4C3 substance:
n (Al4C3) = m (Al4C3) / M (Al4C3) = 86.4 / 144 = 0.6 mol.
The resulting salt from the reaction is aluminum chloride. In the reaction equation, Al4C3 has a coefficient of 1, and AlCl3 has a coefficient of 4. Then the amount of aluminum chloride substance through the amount of aluminum carbide substance is expressed as follows:
n (AlCl3) = n (Al4C3) ⋅ 4/1 = 0.6 ⋅ 4/1 = 2.4 mol.
m (AlCl3) = n (AlCl3) ⋅ M (AlCl3) = 2.4 ⋅ 133.5 = 320.4 g.
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