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

P + 5HNO₃ ⟶ H₃PO₄+ 5NO₂ + H₂O

If 6.2 g of phosphorus was used in the reaction calculate:

(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.

(b) mass of nitric acid consumed at the same time?

(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273°C?

$\begin{matrix} \text{P}& + &5\text{HNO}$3 & \longrightarrow & \text{H}3\text{PO}4 & + & 5\text{NO}2& + &\text{H}_2\text{O} \ 31 \text{g} && 5[1 + 14 + 3(16)] && 3(1) + 31 + 4(16) \ && 315 \text{ g} && 98 \text{ g} \end{matrix}

(a) 31 g of P = 1 mole

∴ 6.2 g of P = $\dfrac{1}{31}$ x 6.2 = 0.2 mole of P

Mass of phosphoric acid formed = ?

31 g of P produces 98 g of phosphoric acid

∴ 6.2 g of P will form $\dfrac{98}{31}$ x 6.2 = 19.6 g

(b) 31 g P reacts with 315 g HNO₃

∴ 6.2 g P will react with $\dfrac{315}{31}$ x 6.2 = 63 g HNO₃

(c) 31 g P produces = 1 mole steam

∴ 6.2 g P produces $\dfrac{1}{31}$ x 6.2 = 0.2 moles

Volume of steam produced at STP = 0.2 × 22.4 = 4.48 litre

V₁ = 4.48 litre

T₁ = 273 K

P₁ = 760 mm Hg pressure

T₂ = 273 + 273 = 546 K

P₂ = 760 mm Hg pressure

V₂ = ?

Using formula:

$\dfrac{\text{P}$1\text{V}1}{\text{T}1} = $\dfrac{\text{P}2\text{V}$2}{\text{T}2}

Substituting in the formula,

$\dfrac{760\times 4.48}{273}$ = $\dfrac{760 \times\text{V}_2}{546}$

V₂ = 2 x 4.48 = 8.96 L

Hence, volume of steam produced = 8.96 L