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ICSE Class 10 Chemistry Question 29 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 29

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Question

Question 28

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?

Answer

P+5HNO3⟶H3PO4+5NO2+H2O31g5[1+14+3(16)]3(1)+31+4(16)315 g98 g\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 = 131\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 9831\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 31531\dfrac{315}{31} x 6.2 = 63 g HNO3

(c) 31 g P produces = 1 mole steam

∴ 6.2 g P produces 131\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:

P1V1T1\dfrac{\text{P}_1\text{V}_1}{\text{T}_1} = P2V2T2\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}

Substituting in the formula,

760×4.48273\dfrac{760\times 4.48}{273} = 760×V2546\dfrac{760 \times\text{V}_2}{546}

V2 = 2 x 4.48 = 8.96 L

Hence, volume of steam produced = 8.96 L