Question

(c) Ammonia burns in oxygen and the combustion, in the presence of a catalyst, may be represented by
$2{\mathrm{NH}}_3+2\frac{1}{2}{\mathrm{O}}_2\stackrel{}{\to }2\mathrm{NO}+3{\mathrm{H}}_2\mathrm{O}$
(i) What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?
(ii) What volume of oxygen, at STP, is required to form 10 moles of product?
[H = 1; N = 14; O = 16. 1 mole of gas occupies 22.4 dm³ (22.4 litre) at STP]

Answer 1

(i) For two moles of nitrogen monoxide formed, three moles of steam is formed. Therefore, we need to first find the number of moles in 1.5 g of NO.

Mass of one mole of NO = (14 + 16) g = 30 g
Number of moles in 1.5 g of NO = $\left(\frac{1.5}{30}\right)=0.05\mathrm{moles}$
Number of moles of steam formed with 2 moles of NO = 3 moles
Number of moles of steam formed along with 0.05 moles of NO = $\left(\frac{0.05\times 3}{2}\right)=0.075\mathrm{moles}$
Mass of one mole of water = {(2 $\times$1) + 16} g = 18 g
Mass of 0.075 moles of water = (18 $\times$ 0.075) g = 1.35 g

(ii) 10 moles of product will include both NO and steam. From the balanced chemical equation we know that:
Moles of oxygen required to form 5 moles of product = 2.5 moles
Moles of oxygen required to form 10 moles of product = $\left(\frac{10\times 2.5}{5}\right)\mathrm{moles}=5\mathrm{moles}$
Therefore, volume of 5 moles of O₂ = (5 $\times$ 22.4) L = 112 L