Question

The chief ore of $Zn$ is its sulphide, $ZnS$. The ore is concentrated by froth floatation process and then heated in air to convert $ZnS$ to $ZnO$. The following reactions take place in the extraction of $Zn$:
$2ZnS+3O_2\to 2ZnO+2S{O}_2\text{(Yield: 80 \%)}$
$ZnO+H_{2}S{O}_4\to ZnS{O}_4+H_{2}O\text{(Yield: 100 \%)}$
$2ZnS{O}_4+2H_{2}O\to 2Zn+2H_{2}S{O}_4+O_2\text{(Yield: 80\%)}$
The number of moles of $ZnS$ required for producing 2 mol of $Zn$ will be:

• A. 3.125
• B. 2.5
• C. 2.125
• D. 4.25

Answer 1

The correct option is A 3.125

$2ZnS+3O_2\to 2ZnO+2S{O}_2\text{(Yield: 80 \%) (i)}$
$ZnO+H_{2}S{O}_4\to ZnS{O}_4+H_{2}O\text{(Yield: 100 \%) (ii)}$
$2ZnS{O}_4+2H_{2}O\to 2Zn+2H_{2}S{O}_4+O_2\text{(Yield: 80 \%) (iii)}$
In reaction (iii),
2 mol of $ZnS{O}_4$ produces 2 mol of $Zn$.
$\text{\% yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$
Theoretical moles of $Zn$ $\frac{2}{80}\times 100=2.5\text{mol}$ = actual moles of $ZnS{O}_4$
In reaction (ii),
2.5 mol will be produced by 2.5 mol of $ZnO$.
Since yield is 100 %, mol of $ZnO$ is 2.5 mol.
In reaction (i),
Theoretical yield of $ZnO$ $=\frac{2.5}{80}\times 100=3.125$ mol of $ZnS$ taken
Therefore, 3.125 mol of $ZnS$ is required for producing 2 mol of $Zn$.