Concentrated nitric acid oxidises phosphorus to phosphoric acid as

P + HNO₃ ⟶H₃PO₄ + H₂O + NO₂

Balance the equation and answer the following questions based on it.

(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 would be the volume of steam produced at the same time if measured at 760 mm. Hg pressure and 273°C.

(Atomic wt of H = 1, N = 14, O = 16, P = 31) ?

Balanced equation is :

P + 5HNO₃ ⟶ H₃PO₄ + H₂O + 5NO₂

(i) Gram molecular mass of phosphorus = 31 g

gram molecular mass of H₃PO₄ = 3(1) + 31 + 4(16) = 3 + 31 + 64 = 98 g

31 g of P makes 98 g of phosphoric acid

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

Hence, mass of phosphoric acid prepared = 19.6 g

(ii) Gram molecular mass of nitric acid (HNO₃) = 1 + 14 + 3(16) = 63 g

31 g of P needs 63 g of nitric acid

∴ 6.2 g of P will need $\dfrac{63}{31}$ x 6.2 = 12.6 g of nitric acid

Hence, mass of nitric acid needed = 12.6 g

(iii) 31 g of P gives 22.4 lit of steam

∴ 6.2 g of P will give $\dfrac{22.4}{31}$ x 6.2 = 4.48 L

Hence, volume of steam produced = 4.48 L