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?

Question

#5"Fe"(s) + 2"KMnO"_4(aq) + 16 "HCl"(aq) -> 5"FeCl"_2(aq) + 2"MnCl"_2(aq) + 2"KCl"(aq) + 8"H"_2"O"(l)#

Serenity Allen · Accepted Answer

#V_(HCl) = "5.59 L"#

Despite its apparent size, the reaction is exactly the same as any other; the balanced reaction (which it is) yields a mol to mol ratio of one reactant to any other reactant or product.

Molarity times volume equals mols because concentration in molarity is equal to mols over volume:

#cancel"L" xx "mol"/cancel"L" = "mol"#

and so, if you can find the mols of #"HCl"#, you can find its volume. To find the mols of #"HCl"#, you need the mols of the limiting reagent.

You assume that the #HCl# completely reacts, so it is not the limiting reagent.

Therefore, it's between the #"Fe"# and the #"KMnO"_4#. I would find the limiting reagent first, so first I would solve for the mols of #"Fe"# and compare it to the mols of #"KMnO"_4#.

#100.0 cancel"g Fe" xx "1 mol"/(55.845 cancel"g Fe") = "1.791 mols Fe"#

#2.00 cancel"L" xx "0.500 mols KMnO"_4/cancel"L" = "1.00 mol KMnO"_4#

Based on the response

#5"Fe"(s) + 2"KMnO"_4(aq) + 16 "HCl"(aq) -> 5"FeCl"_2(aq) + 2"MnCl"_2(aq) + 2"KCl"(aq) + 8"H"_2"O"(l),#

the required stoichiometry is #"5 mols Fe"# to #"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 #"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 #"HCl"# needed based on the limiting reagent:

#1.791 cancel"mols Fe" xx "16 mols HCl"/(5 cancel"mols Fe") = "5.730 mols HCl"#

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

These fractions have to be the same because concentration is an intensive property:

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

Solve for the volume to get #color(blue)(V = "5.59 L")#.

Luna Chen · Answer

undefined To solve this problem, we'll use the balanced chemical equation:

5 Fe(s) + 8 HCl(aq) + 2 KMnO4(aq) → 5 FeCl2(aq) + 2 MnCl2(aq) + 2 KCl(aq) + 8 H2O(l)

From the stoichiometry of the reaction, we see that 8 moles of HCl are required for every 2 moles of KMnO4. We can use this ratio to find the volume of HCl needed.

First, find the number of moles of KMnO4 used:
n(KMnO4) = Molarity(KMnO4) × Volume(KMnO4)
n(KMnO4) = 0.500 mol/L × 2.00 L = 1.00 moles

Since 2 moles of KMnO4 react with 8 moles of HCl, the number of moles of HCl required is:
n(HCl) = (8 moles HCl / 2 moles KMnO4) × n(KMnO4) = 4.00 moles

Now, we can find the volume of HCl needed using its molarity:
Volume(HCl) = n(HCl) / Molarity(HCl) = 4.00 moles / 1.025 mol/L = 3.90 L

So, 3.90 liters of HCl are needed for the reaction.