Question

2.32 g mixture of $FeO$ and $F{e}_2{O}_3$ is burnt in the atmosphere of oxygen. 56 ml of $O_2$ is required at STP for complete oxidation. Mole percentage of $F{e}_2{O}_3$ in the given mixture is:

• A. 50%
• B. 40%
• C. 45%
• D. 60%

Answer 1

The correct option is A 50%

$2FeO+\frac{1}{2}{O}_2⟶F{e}_2{O}_3$

Moles of $O_2=\frac{56}{22.4\times 1000}=2.5$ $millimoles$ $(1$ $mole$ gas occupy $22.4$ $L$ at $STP)$

Moles of $FeO=2.5\times 4=10$ $millimoles$

Weight of $FeO={10}^{-2}\times 72=0.72$ $gm$

Weight of $F{e}_2{O}_3=2.32=0.72=1.60$ $gm$

Moles of $F{e}_2{O}_3=\frac{1.60}{160}=0.01$ $moles$

Moles percentage of $F{e}_2{O}_3=\frac{0.01}{0.02}\times 100=50\%$