5.5 g of a mixture of $\ce{FeSO4.7H2O}$ and $\ce{Fe2(SO4)3.9H2O}$ requires $5.4 \:\mathrm{mL}$ of $0.1 \:\mathrm{N}$ $\ce{KMnO4}$ solution for complete oxidation. Calculate the number of moles of hydrated Ferric Sulphate in the mixture.

My progress towards solving this: I know that I will have to calculate equivalents here, but I'm not sure how to do so using moles. Any hints will do.

Please just suggest one hint ! :)

  • 1
    $\begingroup$ The quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’. $\endgroup$ – user7951 May 12 '15 at 17:35

Find out the n-factors (number of electrons released/required in a half-reaction) of each compound for a full oxidation with $\ce{KMnO4}$. $$\ce{FeSO4 -> Fe^{3+} + SO4^{2-} + e-}$$ (n factor 1)

Ferric sulphate is already oxidised to its maximum, so it doesn't participate here (other than to add weight).

Now, number of equivalents of $\ce{FeSO4}$ is the number of moles times n factor.

You probably can proceed from here :)

Note that if the ferric sulphate was replaced with something like $\ce{As2(SO4)3}$, which follows the reaction $$\ce{As2(SO4)3 -> 2As^{5+} + 3SO4^{2-} + 4e^-}$$

In which case the n factor is 4. Now, you'll have to calculate number of equivalents of $\ce{As2(SO4)3}$ (with an unknown placeholder mass), and solve a pair of simultaneous equations (total mass and law of equivalents) to get the ratios.

  • $\begingroup$ Thanks !! Did this question both by mole equation and equivalents method. $\endgroup$ – DarkMagician Mar 20 '13 at 3:20
  • $\begingroup$ @DarkMagician: That's great :) The concept of equivalents gets confusing when you start redox reactions, but generally you can manage with practice. You say you did it via the mole method -- for this you'd need to know the n-factor of KMnO4 (It's 5, it gets reduced to $\ce{Mn^{2+}}$). Did you already know this, or did you circumvent it? (Also, you can mark this answer as "accepted" by clicking the green tick next to it) $\endgroup$ – ManishEarth Mar 20 '13 at 7:21
  • $\begingroup$ Manish, The only thing which was confusing me was the ionic equation. Mole method requires balancing here, equivalent method does not even require balancing of chemical equation and is far more easy. I am an Indian by the way, and this question was from RC Mukherjee book. :) $\endgroup$ – DarkMagician Mar 20 '13 at 11:12
  • $\begingroup$ @DarkMagician: Ah, got it :) (I can tell your location, as a moderator I can see these things). Never used RC Mukherjee myself (too lazy to solve from that), but it's not bad :) $\endgroup$ – ManishEarth Mar 20 '13 at 11:41

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