25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculate the percentage of fes047h20 in the sample
the right answer 88.96%
My calculation
0.02L x 0.1 = 0.002
0.002 x 1000 /25 =0.08 x 293g/mol /25g x 100% = 93.76%
To calculate the percentage of ferrous sulphate heptahydrate (FeSO4·7H2O) in the sample, you need to use the stoichiometry of the reaction between FeSO4·7H2O and KMnO4.
Here's how you can calculate it step by step:
1. Determine the balanced equation for the reaction between FeSO4·7H2O and KMnO4, which is:
5 FeSO4·7H2O + 2 KMnO4 + 8 H2SO4 → 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O
2. Use the balanced equation to determine the number of moles of KMnO4 reacted.
From the balanced equation, you can see that 2 moles of KMnO4 react with 5 moles of FeSO4·7H2O. You used 20 ml of N/10 KMnO4, which is 0.02 L x 0.1 M, to react with the FeSO4·7H2O.
Moles of KMnO4 used = (0.02 L) x (0.1 mol/L) = 0.002 mol
3. Use the stoichiometry of the reaction to determine the number of moles of FeSO4·7H2O present in the sample.
From the balanced equation, you can see that 5 moles of FeSO4·7H2O react with 2 moles of KMnO4. So, the ratio of moles of FeSO4·7H2O to moles of KMnO4 is 5:2.
Moles of FeSO4·7H2O = (0.002 mol KMnO4) x (5 mol FeSO4·7H2O / 2 mol KMnO4) = 0.005 mol
4. Calculate the molar mass of FeSO4·7H2O.
Molar mass of FeSO4·7H2O = (1 mol of Fe x atomic mass of Fe) + (1 mol of S x atomic mass of S) + (4 mol of O x atomic mass of O) + (14 mol of H x atomic mass of H2O)
= (1 x 55.845) + (1 x 32.06) + (4 x 16.00) + (14 x (1.008 x 2 + 16.00))
= 278.01 g/mol
5. Calculate the mass of FeSO4·7H2O in the sample.
Mass of FeSO4·7H2O = (0.005 mol) x (278.01 g/mol) = 1.39 g
6. Calculate the percentage of FeSO4·7H2O in the sample.
Percentage of FeSO4·7H2O = (1.39 g / 25 g) x 100% = 5.56%
It seems there was an error in your calculation. The correct answer for the percentage of FeSO4·7H2O in the sample is 5.56%, not 93.76%.
To calculate the percentage of FeSO4·7H2O in the sample, you need to use the balanced chemical equation for the reaction between FeSO4·7H2O and KMnO4.
FeSO4·7H2O + 8H2SO4 + 5KMnO4 → K2SO4 + 8H2O + 5MnSO4 + Fe2(SO4)3
From the reaction equation, we can see that 5 moles of KMnO4 react with 1 mole of FeSO4·7H2O.
Given that 25mL of the FeSO4 solution required 20mL of N/10 KMnO4 solution for complete oxidation, we can calculate the equivalent moles of KMnO4 used.
1 L of solution = 1000 mL of solution
So, 25 mL of the FeSO4 solution is equal to 25/1000 = 0.025 L.
Using the equation N1V1 = N2V2, where N1 is the normality of KMnO4, V1 is the volume of KMnO4 used (20 mL or 0.02 L), N2 is the unknown normality of the FeSO4 solution, and V2 is the volume of FeSO4 solution used (0.025 L), we can find N2.
N1V1 = N2V2
N2 = (N1V1) / V2
Since the KMnO4 solution is N/10, N1 = 10/100 = 0.1.
N2 = (0.1 x 0.02) / 0.025
= 0.08M
Now, we can calculate the moles of FeSO4·7H2O using the derived normality.
Number of moles of FeSO4·7H2O = Normality x Volume
= 0.08 x 0.025
= 0.002 moles
The molar mass of FeSO4·7H2O is 278 g/mol.
Mass of FeSO4·7H2O = Number of moles x Molar mass
= 0.002 x 278
= 0.556 g
Finally, we can calculate the percentage of FeSO4·7H2O in the sample.
Percentage = (Mass of FeSO4·7H2O / Mass of sample) x 100
= (0.556 / 25) x 100
= 2.224%
Therefore, the percentage of FeSO4·7H2O in the sample is 2.224%.