Question

Copper on reacting with concentrated H₂SO₄ produces copper sulphate. If 1.28g of copper is to be converted to copper sulphate, find:

(a) The weight of copper sulphate formed.

(b) Weight of sulphuric acid required.

[Cu=64, S=32, O=16]

Answer 1

• Reaction between copper and sulphuric acid :
  $\mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{H}}_2{\mathrm{So}}_4\left(\mathrm{aq}\right)\to {\mathrm{CuSO}}_4\left(\mathrm{aq}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

1mole 2mole(conc.) 1mole

64g 2×98 160g

=196g

(a) $64g$of Cu produces CuSO₄= 160g

• 1.28g of Cu produce CuSO₄ =$\frac{160}{64}\times 1.28$

= 3.2g

• The weight of copper sulphate formed is 3.2g.

(b) 64g of Cu requires H₂SO₄=196g

• 1.28 of Cu requires H₂SO₄ =$$\begin{gathered} \frac{196}{64}\times \frac{1.28}{1} \\ =3.92g \end{gathered}$$
• Therefore, 3.92g of sulphuric acid is required.