X g of KMnO₄ is required to oxidise 0.56 g of FeSO₄ in acidic medium — JEE Chemistry
$x$ g of $KMnO_4$ is required to oxidise $0.56$ g of $FeSO_4$ in acidic medium. Find $x$.
(Molar masses: $KMnO_4 = 158$, $FeSO_4 = 152$ g/mol)
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2mo ago
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The balanced equation in acidic medium:
$$2KMnO_4 + 10FeSO_4 + 8H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O$$
Moles of $FeSO_4 = \dfrac{0.56}{152} = 3.68 \times 10^{-3}$ mol
From ratio $2:10$:
$$Moles of KMnO_4 = \frac{2}{10} \times 3.68\times10^{-3} = 7.37\times10^{-4} mol$$
$$x = 7.37\times10^{-4} \times 158 = 0.116 g$$
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Discussion (4)
D
Can someone explain why we ignore the other root here?
P
Adding for context: NCERT covers the base concept in the same chapter.
NR
Plotting it roughly also helps you sanity-check the sign.
C
Underrated solution. The way you set it up makes it almost obvious.