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"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)#
1 Answer
#V_(HCl) = "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"L" xx "mol"/cancel"L" = "mol"#
and so, if you can find the mols of
You assume that the
Therefore, it's between the
#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#
From the reaction
#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
#"1.791 mols Fe"/("1.00 mol KMnO"_4) < "5 mols Fe"/("2 mols KMnO"_4)#
This means that
Therefore, we can then convert to the exact mols of
#1.791 cancel"mols Fe" xx "16 mols HCl"/(5 cancel"mols Fe") = "5.730 mols HCl"#
So, we need
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