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Question
Aluminium carbide reacts with water according to the following equation.
\[\ce{Al4C3 + 12H2O -> 3CH4 + 4Al(OH)3}\]
What volume of methane is obtained from 12 g of aluminium carbide?
Solution
\[\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}\]
144 g 3 x 22.4 L
67.2 L
144 g of Al4C3 will produce 67.2 L of methane
∴ 12 g of Al4C3 will produce
= `67.2/144 xx 12`
= 5.6 L
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