P + 5HNO3 [conc.] ⟶ H3PO4 + H2O + 5NO2.
If 9.3 g of phosphorous was used in the reaction, calculate :
(i) Number of moles of phosphorous taken.
(ii) The mass of phosphoric acid formed.
(iii) The volume of NO2 produced at s.t.p.
[H = 1, N = 14, P = 31, O = 16]

P + 5HNO₃ → H₃PO₄ + H₂O + 5NO₂

31 g                    98 g

(i) 31 g of P = 1 mole

∴ 9.3 g of P = 1\over 31 × 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 98\over 31 × 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 =5 × 22.4\over 31 × 9.3

= 33.6 lit.