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Find out the amount of $\ce{KMnO4}$ required for the complete oxidation of $\pu{5 mol}$ of an equimolar mixture of ferric oxalate and ferrous oxalate.
I understand that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$ and ferrous ions to ferric ions. I wrote the following reaction: $$\ce{FeC2O4 + Fe2(C2O4)3 + MnO4^- + 8H+ + e- -> 2Fe^3+ + 4CO2 + Mn^2+ + 4H2O}$$
Then, I applied equivalence concept but could not get a correct answer. Any ideas?
I'd like the question to be tackled in two parts. In each part, we're going to deal with one of ferric oxalate and ferrous oxalate.
Since the mixture ($\pu{5 mol}$) given to us is equimolar, we have $\pu{2.5 mol}$ of ferric oxalate and $\pu{2.5 mol}$ of ferrous oxalate. I second your understanding that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$, because the medium is acidic, and ferrous ions to ferric ions.
Ferric oxalate doesn't contain any $\ce{Fe^3+}$ ions, so our main focus here is the conversion of oxalate ($\ce{C2O4^2-}$) into carbon dioxide. The oxidation state of carbon in oxalate is $+3$, while it's $+4$ in carbon dioxide. So there's a net change of $+1$ per carbon atom. Oxalate contains two carbon atoms, so that gives us a valency factor of $2$ per mole of oxalate ion. However, for $1$ mole of $\ce{Fe2(C2O4)3}$, you'll have $3$ moles of oxalate ions that are undergoing the aforementioned change. This gives us an overall valency factor for $\ce{Fe2(C2O4)3}$ as $2 \times 3 = 6$. The valency factor for the $\ce{MnO4- -> Mn^2+}$ change is $5$. Now, we're going to equate the equivalents as per the Law of Equivalence.
$$ \begin{align*} n_{\ce{KMnO_4}} \times V_{f_1} &= n_{\ce{Fe2(C2O4)3}} \times V_{f_1'}\\ x \times 5 &= 2.5 \times 6 \\ x &= 3 \end{align*} $$
For the part of the mixture that contains $\ce{FeC2O4}$, we need to change our valency factor. This is because here we're taking into account the conversion of ferrous ions into ferric ions; this conversion has a valency factor of $1$. So the overall valency factor, in this case, would be $2+1=3$. Now, since this has been done, we proceed to apply the same equation as in the previous case.
$$ \begin{align*} n_{\ce{KMnO_4}} \times V_{f_2} &= n_{\ce{FeC2O4}} \times V_{f_2'}\\ x \times 5 &= 2.5 \times 3 \\ x &= 1.5 \end{align*} $$
Thus, overall, it can be concluded that you'll need $\pu{3+1.5=4.5 mol}$ of $\ce{KMnO4}$ for the complete oxidation. The beauty of this? I didn't use any reaction-equilibrium equations.
This is the reason we use equivalent concept, to save ourselves from the hassle of writing (and perhaps balancing) chemical reactions.
