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Chemistry

66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate:

(a) Mass of ammonia required.

(b) The volume of the gas used at S.T.P.

(c) The mass of acid required.

Mole Concept

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Answer

(a) 2NH3+H2SO4(NH4)2SO42[14+3(1)]2(1)+32+4(16)2[14+4(1)]+32+4(16)=(2×17)=2+32+64=36+32+64=34 g=98 g=132 g2 mole\begin{matrix} 2\text{NH}3 & + &\text{H}2\text{SO}4 & \longrightarrow & \text{(NH}4)2\text{SO}4 \ 2[14 + 3(1)] & & 2(1) + 32 + 4(16) & & 2[14 + 4(1)] + 32 + 4(16) \ = (2 \times 17) & & = 2 + 32 + 64 & & = 36 + 32 + 64 \ = 34 \text{ g} & & = 98 \text{ g} & & = 132 \text{ g} \ 2\text{ mole} \end{matrix}

132 g ammonium sulphate is produced by 34 g of NH3

∴66 g ammonium sulphate is produced by 34132\dfrac{34}{132} x 66 = 17 g of NH3

Hence, 17g of NH3 is required.

(b) 132 g ammonium sulphate uses 2 x 22.4 L of gas

∴ 66 g of ammonium sulphate will use 2×22.4132\dfrac{2 \times 22.4}{132} x 66 = 22.4 litres

(c) For 132 g ammonium sulphate 98 g of acid is required

∴ For 66 g ammonium sulphate 98132\dfrac{98}{132} x 66 = 49 g

Hence, 49g of acid is required.

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