Question

What mass of Zn will be obtained from an ore containing 225 kg of ZnS? Efficiencies of the process have been indicated above the arrow mark.

[Zn = 65, S = 32, O = 16, H = 1]

• A. 134 kg
• B. 112 kg
• C. 102 kg
• D. 130 kg

Answer 1

The correct option is A 134 kg

$\text{moles of}ZnS\text{given}=\frac{225\times 1000}{97.47}mol.\approx 2308.3moles\text{.}$

$\begin{array}{c}\text{by the stoichiometric concepts,}\\ 2\text{amount}2n\text{s sill produce}2\text{amount}ZnO\text{,}\\ \text{but efficiency sould be}90.6\%\end{array}$

$\text{Thus, produced amount of}zn0=2308\cdot 3\times \frac{90\cdot 6}{100}=2091\cdot 327$

$ZnO+H_2{SO}_4\stackrel{100\%}{⟶}{ZnSO}_4+H_{2}O$

$\begin{array}{c}\text{Now, we found out the produced amount of}ZnO=2091.327\text{moles}\\ \text{By stoichiometry, since}1\text{mol of}ZnO\text{produce}1\text{mol of}ZnS{O}_4\\ \text{So, produced amount of}ZnS{O}_4=2091.327\text{molal}\end{array}$

Then,

Similarly, Produced amount of $Zn=2091.327\times \frac{98\cdot 2}{100}$ moles $\therefore$ mass of $\sim 134\cdot 3kg$