Question

Calculate the mass of a coal sample in kg containing 0.05 % mass of iron pyrite $\left(Fe{S}_2\right)$ that can produce 44.8 litres of $S{O}_2$ at 1 atm and $273\text{K}$ with 40 % reaction yield.
$Fe{S}_2+O_2\to F{e}_2{O}_3+S{O}_2$
(Molar mass of Iron = $56\text{g/mol}$; molar mass of Sulphur = $32\text{g/mol}$)

• A. $600\text{kg}$
• B. $550\text{kg}$
• C. $400\text{kg}$
• D. $350\text{kg}$

Answer 1

The correct option is A $600\text{kg}$

$4Fe{S}_2+11O_2\to 2F{e}_2{O}_3+8S{O}_2$
$(8\times 22.4)\text{L}$ of $S{O}_2$ at STP is produced by 4 mol of $Fe{S}_2$
$44.8\text{L}$ will be produced by $=\frac{4}{8\times 22.4}\times 44.8=1\text{mol}Fe{S}_2$
Mass of $Fe{S}_2$ = number of moles $\times$ molar mass $=1\times 120=120\text{g}$
Reaction yield is 40%
So, actual amount of $Fe{S}_2$ required $=\frac{120}{0.4}=300\text{g}$
Mass of iron pyrite in coal = 0.05 %
Mass of coal required $=\frac{300\times 100}{0.05}=600\text{kg}$