MCQ
$2 \,mol$ of $FeSO_4$ are oxidised by $'X'\, mol$ of $KMnO_4$ whereas $2\, mol$ of $FeC_2O_4$ are oxidised by $'Y'\, mol$ of $KMnO_4$. The ratio of $X$ and $Y$ is
- ✓$1 : 3$
- B$1 : 2$
- C$1 : 4$
- D$1 : 5$
✓
Answer
Correct option: A.
$1 : 3$
a
Meq of $\mathrm{FeSO}_{4}=$ Meq of $\mathrm{KMnO}_{4}$
Meq of $\mathrm{FeSO}_{4}=$ Meq of $\mathrm{KMnO}_{4}$
$\begin{array}{l}
{2 \times 1=X \times 5} \\
{X=\frac{2}{5}}
\end{array}$
Meq of $\mathrm{FeC}_{2} \mathrm{O}_{4}=\mathrm{Meq}$ of $\mathrm{KMnO}_{4}$
$2 \times 3=Y \times 5$
$\mathrm{Y}=\frac{6}{5}$
$\therefore \frac{\mathrm{X}}{\mathrm{Y}}=\frac{2 / 5}{6 / 5}=1: 3$
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