<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(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\times 0.020L)=0.00762mol](https://tex.z-dn.net/?f=0.381M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20ferrous%20sulfate%7D%7D%7B0.020L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20ferrous%20sulfate%7D%3D%280.381mol%2FL%5Ctimes%200.020L%29%3D0.00762mol)
For the given chemical equation:
![5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O](https://tex.z-dn.net/?f=5Fe%5E%7B2%2B%7D%2B8H%5E%2B%2BMnO_4%5E-%5Crightarrow%205Fe%5E%7B3%2B%7D%2BMn%5E%7B2%2B%7D%2B4H_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 =
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)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL](https://tex.z-dn.net/?f=0.152mol%2FL%3D%5Cfrac%7B0.00152mol%7D%7B%5Ctext%7BVolume%20of%20permanganate%20ion%20%28potassium%20permanganate%29%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20permanganate%20ion%20%28potassium%20permanganate%29%7D%3D%5Cfrac%7B0.00152mol%7D%7B0.152mol%2FL%7D%3D0.01L%3D10.0mL)
Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL