Question

What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following net ionic equation? 5 Fe2+ + 8 H+ + MnO4- → 5 Fe3+ + Mn2+ + 4 H2O

Answer 1

Answer: The volume of permanganate ion (potassium permanganate) is 10.0 mL

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(1)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

`0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L* 0.020L)=0.00762mol`

For the given chemical equation:

`5Fe^(2+)+8H^++MnO_4^-\rightarrow 5Fe^(3+)+Mn^(2+)+4H_2O`

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with =
`(1)/(5)* 0.00762=0.00152mol` of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

`0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=(0.00152mol)/(0.152mol/L)=0.01L=10.0mL`

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL