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Chemistry

Aluminium carbide reacts with water according to the following equation :

Al4C3 + 12H2O ⟶ 4Al(OH)3 + 3CH4

(i) State what mass of aluminium hydroxide is formed from 12 g. of aluminium carbide.

(ii) State the volume of methane at s.t.p. obtained from 12 g of aluminium carbide.

[relative molecular weight of Al4Cl3] = 144 ; Al(OH)3 = 78

Stoichiometry

ICSE 2018

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Answer

Al4C3+12H2O⟶4Al(OH)3+3CH4144 g4(78)3(22.4)=312 g=67.2 lit.\begin{matrix} \text{Al}4\text{C}3 & + & 12\text{H}2\text{O} & \longrightarrow & 4\text{Al(OH)}3 & + & 3\text{CH}_4 \ 144 \text{ g} & & & & 4(78) & & 3(22.4) \ & & & & = 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 312144\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 67.2144\dfrac{67.2}{144} x 12 = 5.6 lit.

Hence, vol. of methane obtained at s.t.p. = 5.6 lit.

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