Question

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

\[\ce{P + 5HNO3_{(conc.)}-> H3PO4 + H2O + 5NO2}\]

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

1. Number of moles of phosphorus taken.
2. The mass of phosphoric acid formed.
3. The volume of nitrogen dioxide produced at STP.

Answer 1

Given: \[\ce{P + 5HNO3_{(conc.)}-> H3PO4 + H2O + 5NO2}\]

(i) Number of moles of phosphorous taken = `9.3/31` = 0.3 mole

Mass of H₃PO₄ = (3 × 1) + 31 + 4(16)

= 98 g

(ii) 1 mole of phosphorous gives = 98 gm of phosphoric acid

∴ 0.3 mole of phosphorous gives = (0.3 × 98 gm) of phosphoric acid

= 29.4 gm of phosphoric acid

(iii) 1 mole of phosphoric gives = 112 L of NO₂ gas at STP

∴ 0.3 moles of phosphorous gives = (112 × 0.3)L of NO₂ gas at STP

= 33.6 L of NO₂ gas at STP