Question

(a) Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
P + 5HNO₃ $\stackrel{}{\to }$ H₃PO₄ + H₂O + 5NO₂
(i) What mass of phosphoric acid can be prepared from 6.2 g of phosphorus?
(ii) What mass of nitric acid will be consumed at the same time?
(iii) What will be the volume of steam at the same time measured at 760 mm Hg pressure and 373°C? (H = 1; N = 14; O = 16; P = 31)

(b) Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation:
4NH₃ + 5O₂ $\stackrel{}{\to }$ 4NO + 6H₂O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

Answer 1

(a)
(i) According to the balanced chemical equation, one mole of phosphorus produces one mole of phosphoric acid.
Number of moles of phosphorus in 6.2 g = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{phosphorus}}{\mathrm{Molecular}\mathrm{weight}\mathrm{of}\mathrm{phosphorus}}=\frac{6.2}{31}=0.2\mathrm{mol}$
Hence, 0.2 moles of phosphoric acid will be produced.
Mass of 0.02 moles of phosphoric acid = Number of moles $\times$ Molecular mass of phosphoric acid
= (0.2 $\times$ 98) g = 19.6 g

(ii) According to the equation, one mole of phosphorus is oxidised by five moles of nitric acid.
As 0.2 moles of phosphorus is consumed in the reaction,
Number of moles of nitric acid consumed = (5 $\times$ 0.2) moles = 1 mole
Mass of one mole of nitric acid = 1 $\times$ Molecular mass of nitric acid = (1 $\times$ 63) g = 63 g

(iii) Number of moles of steam produced from 0.2 moles of phosphorus = 0.2 mole
According to the ideal gas equation,
PV = nRT
It is given that,
P = 760 mmHg = 1 atm
n = 0.2 moles
R = 0.0821 L atm K^-1 mol^-1
T = 373^oC = (373 + 273) K = 646 K
Substituting these values in the ideal gas equation, we get:
1 atm $\times$ V = 0.2 mol $\times$ 0.0821 L atm K^-1 mol^-1 $\times$ 646 K
V = 10.61 L
Thus, 10.61 litres of steam is produced from 0.02 moles of phosphorus.

(b) Let number of moles of ammonia be x.
4NH₃ + 5O₂ $\to$ 4NO + 6H₂O
Initial moles x $\frac{5}{4}x$ 0 0
Final moles 0 0 x $\frac{6}{4}x$
Total number of moles of reactants, initially = $x+\frac{5}{4}x=\frac{9}{4}x$
Total volume of reactants = 27 L
According to the ideal gas equation,
PV = nRT
$$\begin{gathered} P\left(27\right)=\left(\frac{9}{4}x\right)\mathrm{R}T \\ x=\frac{12P}{\mathrm{R}T} \end{gathered}$$

Now, according to the balanced chemical equation, number of moles of nitrogen monoxide is same as the number of moles of ammonia, which is x.
Volume of nitrogen monoxide = $\left(\frac{12P}{\mathrm{R}T}\right)\frac{\mathrm{R}T}{P}=12\mathrm{L}$