Chemistry

The relative molecular mass [mol. wt.] of copper oxide is 80. What vol. of NH₃ (measured at s.t.p.) is required to completely reduce 120 g of CuO.
[3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂].

Stoichiometry

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Answer

$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \ 3\text{ mol.} & & 2 \text{ mol.} \ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \ = 240 \text{ g.} & & = 44.8\text{ lit.} \ \end{matrix}$

$\longrightarrow 3\text{Cu} + 3\text{H}$

240 g of CuO is reduced by 44.8 lit of NH₃

∴ 120 g of CuO is reduced by $\dfrac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH₃ is required

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Related Questions

The reaction of potassium permanganate with acidified iron (II) sulphate is given below :
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ ⟶ K₂SO₄ + 2MnSO₄ + 5 Fe₂(SO₄)₃ + 8H₂O.
If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.
[K = 39, Mn = 55, Fe = 56, S = 32, O = 16]
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