Question

The chief ore of $Zn$ is the sulphide, $ZnS$. The ore is concentrated by froth floatation process and then heated in air to convert $ZnS$ to $ZnO$.
$2ZnS+3O_2\to 2ZnO+2S{O}_2$ ($75\%$ conversion)
$ZnO+H_{2}S{O}_4\to ZnS{O}_4+H_{2}O$ ($100\%$ conversion )
$2ZnS{O}_4+2H_{2}O\to 2Zn+2H_{2}S{O}_4+O_2$ ($80\%$ conversion)

The mass of $Zn$ obtained from a sample of ore containing $292.2$ kg of $ZnS$ will be :

• A. $117.72$ kg
• B. $214.62$ kg
• C. $145.23$ kg
• D. $166.55$ kg

Answer 1

Answer: (A)

$117.72$ kg

$2ZnS+3O_2\stackrel{75\%}{\to }2ZnO+2S{O}_2$

$2ZnO+2H_{2}S{O}_4\stackrel{100\%}{\to }2ZnS{O}_4+2H_{2}O$

$2ZnS{O}_4+2H_{2}O\stackrel{80\%}{\to }2Zn+2H_{2}S{O}_4+O_2$
Form the chemical reactions, it can be observed that one mole of $ZnS$ gives $1$ mole of $Zn$.
The molar masses of $Zn$ and $ZnS$ are $65.37$ g/mol and $97.47$ g/mol.
The mass of zinc that can be obtained from $292.2$ kg of $ZnS$ is $\frac{65.37}{97.44}\times 292.2\times \frac{75}{100}\times \frac{100}{100}\times \frac{80}{100}=117.72kg$.