Moles of K_2Cr_2O_7 used to oxidise 1 mole of | vedclass

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(C) In $Fe_{0.92}O$,let the oxidation state of $Fe$ be $x$. Since the oxidation state of $O$ is $-2$,we have $0.92x + (-2) = 0$,so $x = \frac{2}{0.92}

Vedclass · Accepted Answer

$\frac{0.76}{6}$

Vedclass · Answer

(C) In $Fe_{0.92}O$,let the oxidation state of $Fe$ be $x$. Since the oxidation state of $O$ is $-2$,we have $0.92x + (-2) = 0$,so $x = \frac{2}{0.92}$.When $1$ $mole$ of $Fe_{0.92}O$ is oxidized to $Fe^{3+}$,the change in oxidation state of $Fe$ is $(3 - \frac{2}{0.92}) 	imes 0.92 = 3 	imes 0.92 - 2 = 2.76 - 2 = 0.76$.$K_2Cr_2O_7$ acts as an oxidizing agent where $Cr$ changes from $+6$ to $+3$,a change of $3$ per $Cr$ atom,or $6$ per $K_2Cr_2O_7$ molecule.Equating the total change in oxidation states: $n 	imes 6 = 0.76$.Therefore,$n = \frac{0.76}{6}$ $moles$.

Moles of $K_2Cr_2O_7$ used to oxidise $1$ $mole$ of $Fe_{0.92}O$ to $Fe^{3+}$ are -

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