Q18 / 180 · COMEDK 2024 (Memory Based) · Chemistry
How many grams of potassium dichromate are required to oxidise 20.0 g of $\mathrm{Fe}^{2+}$ in $\mathrm{FeSO}_4$ to $\mathrm{Fe}^{3+}$ if the reaction carried out in an acidic medium? Molar mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and $\mathrm{FeSO}_4$ are 294 and 152 respectively.
- Correct answerA6.5g
- B5.5g
- C3.5g
- D4.5g
Solution
$\text { } \begin{aligned}<br/>\text { Number of moles of } \mathrm{FeSO}_4 & =\frac{20 \mathrm{~g}}{152 \mathrm{~g} \mathrm{~mol}^{-1}} \\<br/>& =0.132 \mathrm{~mole}<br/>\end{aligned}$
The balance ionic equation of redox reaction is
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_2 \mathrm{O}$
6 moles of $\mathrm{Fe}^{2+}=1$ mole of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$\therefore 0.132$ mole of $\mathrm{Fe}^{2+}=\frac{1}{6} \times 0.132$
$=0.022 \text { moles of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
Mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in 0.02 moles
$=0.022 \times 294=6.5 \mathrm{~g}$
The balance ionic equation of redox reaction is
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_2 \mathrm{O}$
6 moles of $\mathrm{Fe}^{2+}=1$ mole of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$\therefore 0.132$ mole of $\mathrm{Fe}^{2+}=\frac{1}{6} \times 0.132$
$=0.022 \text { moles of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
Mass of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in 0.02 moles
$=0.022 \times 294=6.5 \mathrm{~g}$
Hence, the correct answer is option: 'A'.