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Chemistry

(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH4Cl by the reaction:

2NH4Cl + Ca(OH)2 ⟶ CaCl2 +2H2O + 2NH3

(b) What will be the volume of ammonia when measured at S.T.P?

Mole Concept

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Answer

2NH4Cl+ Ca(OH)2⟶CaCl2 + 2H2O +2NH32[14+4(1)+35.5]2[14+3(1)]107g34g2 mole\begin{matrix} 2\text{NH}4\text{Cl} & + \text{ Ca(OH)}2 \longrightarrow \text{CaCl}2 \space + \space 2\text{H}2\text{O} \space + & 2\text{NH}_3 \ 2[14 + 4(1) + 35.5] & & 2[14 + 3(1)] \ 107 \text{g} & &34\text{g} \ & &2\text{ mole} \ \end{matrix}

(a) 107 g NH4Cl gives 34 g of NH3

∴ 21.4 g NH4Cl will give 34107\dfrac{34}{107} x 21.4

= 6.8 g of NH3

(b) Volume of ammonia produced by 107 g NH4Cl = 2 x 22.4 L

∴ Volume of ammonia produced by 21.4 g NH4Cl = 2×22.4107\dfrac{2 \times 22.4}{107} x 21.4

= 8.96 L

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