Q.56. In order to oxidise a mixture of 1 mole each of FeCO, Fe(CO), FeSO and Fe(SO), in acidic mediu

Q: What is the correct answer for this exam Chemistry Q56?

The correct answer is B. We need to find the total number of equivalents of reducing agents and equate it to the equivalents of KMnO$_4$.In acidic medium, KMnO$_4$ acts as an oxidizing agent, and Mn changes from +7 to +2, so its n-factor is 5.1. FeC$_2$O$_4$: Fe is +2 (n-factor 1) and C$_2$O$_4^{2-}$ is (2 carbons at +3, each changes to +4, so 2 electrons per C$_2$O$_4^{2-}$, n-factor 2). Total n-factor for FeC$_2$O$_4$ =

Q: Q.56. In order to oxidise a mixture of 1 mole each of FeCO, Fe(CO), FeSO and Fe(SO), in acidic medium, the number of moles of KMnO required is :

The correct answer is B. We need to find the total number of equivalents of reducing agents and equate it to the equivalents of KMnO$_4$.In acidic medium, KMnO$_4$ acts as an oxidizing agent, and Mn changes from +7 to +2, so its n-factor is 5.1. FeC$_2$O$_4$: Fe is +2 (n-factor 1) and C$_2$O$_4^{2-}$ is (2 carbons at +3, each changes to +4, so 2 electrons per C$_2$O$_4^{2-}$, n-factor 2). Total n-factor for FeC$_2$O$_4$ =

Question

Q.56. In order to oxidise a mixture of 1 mole each of FeCO, Fe(CO), FeSO and Fe(SO), in acidic medium, the number of moles of KMnO required is :

FirstInTest · Accepted Answer

We need to find the total number of equivalents of reducing agents and equate it to the equivalents of KMnO$_4$.In acidic medium, KMnO$_4$ acts as an oxidizing agent, and Mn changes from +7 to +2, so its n-factor is 5.1. FeC$_2$O$_4$: Fe is +2 (n-factor 1) and C$_2$O$_4^{2-}$ is (2 carbons at +3, each changes to +4, so 2 electrons per C$_2$O$_4^{2-}$, n-factor 2). Total n-factor for FeC$_2$O$_4$ = 1 + 2 = 3. Moles = 1, Equivalents = $1 	imes 3 = 3$.2. Fe$_2$(C$_2$O$_4$)$_3$: Fe is +3 (n-factor 0 for Fe as it's already in highest oxidation state). C$_2$O$_4^{2-}$ (n-factor 2). There are 3 C$_2$O$_4^{2-}$ ions. Total n-factor for Fe$_2$(C$_2$O$_4$)$_3$ = $3 	imes 2 = 6$. Moles = 1, Equivalents = $1 	imes 6 = 6$.3. FeSO$_4$: Fe is +2 (n-factor 1). SO$_4^{2-}$ is (S at +6, n-factor 0). Total n-factor for FeSO$_4$ = 1. Moles = 1, Equivalents = $1 	imes 1 = 1$.4. Fe$_2$(SO$_4$)$_3$: Fe is +3 (n-factor 0). SO$_4^{2-}$ is (S at +6, n-factor 0). Total n-factor for Fe$_2$(SO$_4$)$_3$ = 0. Moles = 1, Equivalents = $1 	imes 0 = 0$.Total equivalents of reducing agents = $3 + 6 + 1 + 0 = 10$.Let 'x' be the moles of KMnO$_4$. Equivalents of KMnO$_4$ = $x 	imes 5$.Equating equivalents: $5x = 10 \Rightarrow x = 2$.So, 2 moles of KMnO$_4$ are required.

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3

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2

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5

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7

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