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 $Zn$:
$2ZnS+3O_2\to 2ZnO+2S{O}_2\text{(Yield: 62.5 \%)}$
$ZnO+H_{2}S{O}_4\to ZnS{O}_4+H_{2}O\text{(Yield: 80 \%)}$
$2ZnS{O}_4+2H_{2}O\to 2Zn+2H_{2}S{O}_4+O_2\text{(Yield: 80\%)}$
The number of moles of $ZnS$ required for producing $2$ mol of $Zn$ will be:

• A. $4.25\text{mol}$
• B. $3.00\text{mol}$
• C. $6.75\text{mol}$
• D. $5.00\text{mol}$

Answer 1

The correct option is D $5.00\text{mol}$

Given,
$2ZnS+3O_2\to 2ZnO+2S{O}_2\text{(Yield: 60 \%) (i)}$
$ZnO+H_{2}S{O}_4\to ZnS{O}_4+H_{2}O\text{(Yield: 80 \%) (ii)}$
$2ZnS{O}_4+2H_{2}O\to 2Zn+2H_{2}S{O}_4+O_2\text{(Yield: 80 \%) (iii)}$
In reaction (iii),
$2$ mol of $ZnS{O}_4$ produces $2$ mol of $Zn$.
$\text{\% yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$
$\therefore \frac{2}{80}\times 100=2.5\text{mol}=\text{actual moles of}ZnS{O}_4$
In reaction (ii),
$2.5$ mol will be produced from $2.5$ mol of $ZnO$.
Since yield is $80\%$,
$\therefore \frac{2.5}{80}\times 100=3.125\text{mol}=\text{actual moles of}ZnO$
In reaction (i), the yield of the reaction is $62.5\%$
$ZnO$ produced in the reaction is $3.125\text{g}$
$\therefore \frac{3.125}{62.5}\times 100=5$ mol of $ZnS$ taken
Therefore, $5$ mol of $ZnS$ is required for producing $2$ mol of $Zn$.