# For the reaction of "2.00 L" of "0.500 M" potassium permanganate with "100.0 g" of iron solid in the presence of "1.025 M HCl", what volume of "HCl" is needed?

## $5 \text{Fe"(s) + 2"KMnO"_4(aq) + 16 "HCl"(aq) -> 5"FeCl"_2(aq) + 2"MnCl"_2(aq) + 2"KCl"(aq) + 8"H"_2"O} \left(l\right)$

Oct 20, 2017

${V}_{H C l} = \text{5.59 L}$

Even though the reaction looks big, it is just like any other reaction; you get a mol to mol ratio of some reactant to any other reactant or product from the balanced reaction (which it is).

Since concentration in molarity is mols over volume, molarity times volume is mols:

$\cancel{\text{L" xx "mol"/cancel"L" = "mol}}$

and so, if you can find the mols of $\text{HCl}$, you can find its volume. To find the mols of $\text{HCl}$, you need the mols of the limiting reagent.

You assume that the $H C l$ completely reacts, so it is not the limiting reagent.

Therefore, it's between the $\text{Fe}$ and the ${\text{KMnO}}_{4}$. I would find the limiting reagent first, so first I would solve for the mols of $\text{Fe}$ and compare it to the mols of ${\text{KMnO}}_{4}$.

$100.0 \cancel{\text{g Fe" xx "1 mol"/(55.845 cancel"g Fe") = "1.791 mols Fe}}$

$2.00 {\cancel{\text{L" xx "0.500 mols KMnO"_4/cancel"L" = "1.00 mol KMnO}}}_{4}$

From the reaction

$5 \text{Fe"(s) + 2"KMnO"_4(aq) + 16 "HCl"(aq) -> 5"FeCl"_2(aq) + 2"MnCl"_2(aq) + 2"KCl"(aq) + 8"H"_2"O} \left(l\right) ,$

the required stoichiometry is $\text{5 mols Fe}$ to ${\text{2 mols KMnO}}_{4}$. Or, we could write that as a mol to mol ratio. Compare to what you calculated:

"1.791 mols Fe"/("1.00 mol KMnO"_4) < "5 mols Fe"/("2 mols KMnO"_4)

This means that $\text{Fe}$ is our limiting reagent. We have less of it than the reaction asks for.

Therefore, we can then convert to the exact mols of $\text{HCl}$ needed based on the limiting reagent:

$1.791 \cancel{\text{mols Fe" xx "16 mols HCl"/(5 cancel"mols Fe") = "5.730 mols HCl}}$

So, we need $\text{5.730 mols}$ of $\text{HCl}$ to be present within some volume of reagent solution so that its concentration is $\text{1.025 mols HCl"/"L soln}$.

Since concentration is an intensive property, these fractions must be equal:

"1.025 mols HCl"/"L soln" = "5.730 mols"/(?" ""L soln")

Solve for the volume to get $\textcolor{b l u e}{V = \text{5.59 L}}$.