Question

A sample of $1.0$ g solid $F{e}_2{O}_3$ of $80$% purity is dissolved in a moderately concentrated $HCl$ solution which is reduced by $Zn$ dust. The resulting solution required $16.7$ ml of a $0.1$M solution of oxidant. Calculate the number of electrons taken up by the oxidant.

• A. $5$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is D $6$

$1.0$ g of $80$% pure $F{e}_2{O}_3$ contains $\frac{1\times 80}{100}=0.8$ g pure $F{e}_2{O}_3$.

The reactions taking place in this process are as follows:

$F{e}_2{O}_3+6HCl\to 2FeC{l}_3+3H_{2}O$

$2FeC{l}_3+H_2\stackrel{zincdust}{\to }2FeC{l}_2+2HCl$

$F{e}^{2+}+oxidant\to F{e}^{3+}+reductant$

The equivalent weight of $F{e}_2{O}_3$ is $\frac{Mol.wt.}{2}=\frac{160}{2}=80$ g.

The number of milli-equivalents of $F{e}_2{O}_3$ is equal to the number of milli-equivalents of oxidant.

$\frac{0.8}{80}\times 1000=16.7\times 0.1\times x$

Hence, $x=6$, where $x$ is the number of electrons taken up by the oxidant.

Therefore, option D is correct.