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Chemistry

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]

Stoichiometry

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Answer

P+5HNO3⟶H3PO4+H2O+5NO231 g[conc.]3(1)+31+4(16)=98 g\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 = 131\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 9831\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 = 5×22.431\dfrac{5 \times 22.4}{31} x 9.3 = 33.6 lit.

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