WAEC Chemistry 2016 Paper

Concept:Use the mole ratio from the balanced equation to convert the given mass of ethene into the required volume of oxygen at s.t.p.Explanation:From the equation, 1 mole of ethene, $\mathrm{C_2H_4}$, reacts with 3 moles of oxygen, $\mathrm{O_2}$.The molar mass of ethene is 28 g, so 28 g of ethene corresponds to 1 mole.Thus, 28 g of ethene requires $3 \times 22.4 \, \text{dm}^3 = 67.2 \, \text{dm}^3$ of oxygen.For 14 g of ethene, the volume of oxygen needed is:$$\frac{14}{28} \times 67.2 \, \text{dm}^3 = 33.6 \, \text{dm}^3$$Answer:B. $33.6 \, \text{dm}^3$