Mole Concept and Stoichiometry — Question 9

$\begin{matrix} \text{Al}_4\text{C}_3 & + & 12\text{H}_2\text{O} & \longrightarrow & 4\text{Al (OH)}_3 & + & 3\text{CH}_4 \\ 4(27) + 3(12) & & & & 4(78) & & 3(22.4) \\ = 144 \text{ g} & & & & = 312 \text{ g} & & = 67.2 \text{ lit.} \\ \end{matrix}$

144 g of aluminium carbide forms 312 g of aluminium hydroxide.

∴ 12 g of aluminium carbide will form $\dfrac{312}{144}$ x 12 = 26 g of aluminium hydroxide

Hence, 26 g of aluminium hydroxide is formed.

(ii) 144 g of aluminium carbide forms 67.2 lit of methane.

∴ 12 g of aluminium carbide will form $\dfrac{67.2}{144}$ x 12 = 5.6 lit.

Hence, vol. of methane obtained = 5.6 L