Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5HNO₃ [conc.] ⟶ H₃PO₄ + H₂O + 5NO₂.

If 9.3 g of phosphorus was used in the reaction, calculate :

(i) Number of moles of phosphorus taken.

(ii) The mass of phosphoric acid formed.

(iii) The volume of nitrogen dioxide produced at STP.

$\begin{matrix} \text{P} & + & 5\text{HNO}$3 & \longrightarrow & \text{H}3\text{PO}4 & + & \text{H}2\text{O} & + & 5\text{NO}_2 \ 31 \text{ g} & & [conc.] & & 3(1) + 31 \ & & & & + 4(16) \ & & & & = 98 \text{ g} \ \end{matrix}

(i)

31 g of P = 1 mole

∴ 9.3 g of P = $\dfrac{1}{31}$ x 9.3 = 0.3 moles.

Hence, 0.3 moles of phosphorous was taken for the reaction.

(ii)

31 g of P forms 98 g of phosphoric acid

∴ 9.3 g will form $\dfrac{ 98}{31}$ x 9.3 = 29.4 g.

Hence, 29.4 g. of phosphoric acid is formed

(iii)

31 g of P produces 5 vol = 5 x 22.4 lit.

∴ 9.3 g will produce = $\dfrac{5 \times 22.4}{31}$ x 9.3 = 33.6 lit.