Quantity of water in a redox problem


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> $\pu{76 g}$ of $\ce{FeSO4}$ are oxidized by $\pu{0.04 M}$ solution of $\ce{KMnO4}$ in presence of $80\%$ $\ce{H2SO4}$. What quantity of water must be added to the final solution of $\ce{Fe2(SO4)3}$ for the concentration to be $\pu{0.25 M}$?
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I wrote the reaction:


$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$


But I have no idea how to find the amount of water. I found the quantity of $\ce{Fe2(SO4)3}$ to be $\pu{100 g}$, but I have no idea to find all the water.




To achieve the necessary concentration, based on the water presence in the system the following condition must be fulfilled:


$$C(\ce{Fe2(SO4)3}) = \frac{n(\ce{Fe2(SO4)3})}{V\_1 + V\_2 + V\_x} \tag{1}$$


where $C$ – concentration, $n$ – amount, $V\_1$ – volume of water remaining from permanganate solution, $V\_2$ – volume of water synthesized during the reaction, $V\_x$ – additional volume of water to be found.


From the balanced redox reaction, which it seems you have established correctly:


$$\ce{10 FeSO4 + 8 H2SO4 + 2 KMnO4 -> 5 Fe2(SO4)3 + K2SO4 + 2 MnSO4 + 8 H2O}$$


one can express both $V\_1$ and $V\_2$, as well as the amount of iron(III) sulfate using molar ratios between reactants and products and density of water $\rho(\ce{H2O})$:


$$V\_1 = \frac{n(\ce{KMnO4})}{C(\ce{KMnO4})} = \frac{2 \cdot n(\ce{\ce{FeSO4}})}{10 \cdot C(\ce{KMnO4})} \tag{2}$$


$$V\_2 = \frac{m(\ce{H2O})}{\rho (\ce{H2O})} = \frac{8 \cdot n(\ce{FeSO4}) \cdot M(\ce{H2O})}{10 \cdot \rho (\ce{H2O})} \tag{3}$$


$$n(\ce{Fe2(SO4)3}) = \frac{5 \cdot n(\ce{FeSO4})}{10} \tag{4}$$


From (1):


$$V\_x = \frac{n(\ce{Fe2(SO4)3})}{C(\ce{Fe2(SO4)3})} - V\_1 - V\_2 \tag{5}$$


Combining (2–4) into (5), and taking into account that


$$n(\ce{FeSO4}) = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \tag{6}$$


one can obtain the following expression for the volume of water to be added:


$$V\_x = \frac{m(\ce{FeSO4})}{M(\ce{FeSO4})} \left( \frac{1}{2 \cdot C(\ce{Fe2(SO4)3})} - \frac{1}{5 \cdot C(\ce{KMnO4})} - \frac{4 \cdot M(\ce{H2O})}{5 \cdot \rho (\ce{H2O})} \right) \tag{7}$$


The problem is, plugging in the current values returns negative volume, which means the solution is *already* too diluted. That much water could've been introduced only by humongous amount of oxidant required due to low concentration, so I guess that the concentration of $\ce{KMnO4}$ is higher than $\pu{0.04 M}$ and there is a typo. Let's assume $C(\ce{KMnO4}) = \pu{0.4 M}$, then:


$$V\_x = \frac{\pu{76 g}}{\pu{152 g mol-1}} \left( \frac{1}{2 \cdot \pu{0.25 M}} - \frac{1}{5 \cdot \pu{0.4 M}} - \frac{4 \cdot \pu{18 g mol-1}}{5 \cdot \pu{1e3 g L-1}} \right) = \pu{0.743 L}~\text{or}~\pu{743 mL}$$


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