Question

What weight of zinc would be required to produce enough hydrogen to reduce 8.5 g of copper oxide completely into copper?

Answer 1

Solution:-

Reaction of zinc to produce hydrogen-

$Zn+2HCl⟶Zn{Cl}_2+H_2$

Reaction of hydrogen to reduce copper oxide into copper-

$CuO+H_2⟶H_{2}O+Cu$

From the above reactions-

$1$ moles of $Zn$ produces $1$ mole of hydrogen gas and $1$ mole of hydrogen required to reduce $1$ mole of $CuO$.

Mole of $CuO$ reduced $=$ mole of $Zn$ required

Therefore,

No. of moles of $CuO$ reduced $=\frac{\text{moles of}CuO}{\text{molar mass of}CuO}$

Given weight of $CuO=8.5g$

Molecular weight of $CuO=79.5g$

Therefore,

No. of moles of $CuO$ reduced $=\frac{8.5}{79.5}=0.107\text{mol}$

Therefore,

No. of moles of $Zn$ required $=0.107\text{mol}$

Now,

Molecular weight of $Zn=65.4g$

$\therefore$ Weight of $Zn$ required $=0.107\times 65.4=6.99g$

Hence $6.99g$ of $Zn$ will be required.