WAEC Chemistry 2019 Paper

Concept:Use the mole ratio from the balanced equation to convert moles of aluminium tetraoxosulphate(VI) to moles of tetraoxosulphate(VI) acid needed.Explanation:The balanced equation is:$$\mathrm{2Al(OH)_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 6H_2O}$$From the equation, $3$ moles of $\mathrm{H_2SO_4}$ produce $1$ mole of $\mathrm{Al_2(SO_4)_3}$.So the mole ratio is $\mathrm{H_2SO_4 : Al_2(SO_4)_3} = 3 : 1$.Let $y$ be the moles of $\mathrm{H_2SO_4}$ needed to produce $0.5$ mol of $\mathrm{Al_2(SO_4)_3}$.Using proportion:$$y = \frac{3 \times 0.5}{1}$$$$y = 1.5$$Thus, $1.5$ moles of tetraoxosulphate(VI) acid is required.Answer:The correct option is C. 1.5.