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 f2s04 7h20 in the sample

1 g of mixture of na2c03 and k2c03 was mole made upto 250ml is aqueous solution 25ml of this solution was neutralized 20ml of hcl of unknown required 16.24ml of 0.1N agn03 for precipitation. Calculate the % k2c03 is mixture , conc of hcl in g/litre and molarity of hcl.

To calculate the percentage of FeSO4·7H2O in the sample, we need to determine the number of moles of FeSO4·7H2O in the 25g sample and then convert it to a percentage.

1. Calculate the number of moles of FeSO4·7H2O:
- The molar mass of FeSO4·7H2O is (55.85 + 32.06 + 4 * 16 + 7 * (1.01 * 2)) g/mol = 278.01 g/mol
- Number of moles = Mass / Molar mass = 25g / 278.01 g/mol = 0.0899 mol

2. Use the equation balanced for the reaction between FeSO4·7H2O and KMnO4:
5FeSO4 + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O + SO4^2-

From the balanced equation, we can see that it takes 5 moles of FeSO4 to react with 1 mole of KMnO4. Therefore, 0.0899 mol of FeSO4·7H2O will react with:
(0.0899 mol / 5) * (1 mole of KMnO4/1 mole of FeSO4) = 0.01798 mol of KMnO4

3. Now, we can find the concentration of KMnO4 and use it to determine the percentage of FeSO4·7H2O:
- 25 mL of the FeSO4 solution required 20 mL of N/10 KMnO4 for complete oxidation.
- N/10 KMnO4 solution means it has the molarity of (1/10) M.

Using the equation:
M1 * V1 = M2 * V2
where M1 = Molarity of KMnO4, V1 = Volume of KMnO4 solution used, M2 = Molarity of FeSO4 solution, and V2 = Volume of FeSO4 solution used.

(1/10 M) * (20 mL) = M2 * (25 mL)
M2 = (1/10 M) * (20 mL) / (25 mL) = 0.08 M

4. To calculate the percentage of FeSO4·7H2O:
- The molar mass of FeSO4·7H2O is 278.01 g/mol
- The mass of FeSO4·7H2O in the 25 mL solution is:
= 0.08 M * (25 mL / 1000 mL) * (278.01 g/mol) = 0.556 g

Percentage of FeSO4·7H2O = (0.556 g / 25 g) * 100% = 2.22%

Therefore, the percentage of FeSO4·7H2O in the sample is approximately 2.22%.

To calculate the percentage of FeSO4·7H2O in the sample, we need to determine how much FeSO4·7H2O was oxidized by KMnO4. Here's how you can calculate it step by step:

1. Write the balanced equation for the oxidation reaction between FeSO4·7H2O and KMnO4:

5FeSO4·7H2O + 2KMnO4 + 8H2SO4 -> 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

2. Calculate the number of moles of KMnO4 required for the reaction:

Given that 20 mL of N/10 KMnO4 solution is used, the equation tells us that 1 mole of KMnO4 corresponds to 5 moles of FeSO4·7H2O. Thus, the moles of KMnO4 used can be calculated as follows:

Moles of KMnO4 = (20 mL * 0.1 mol/L) / 1000 = 0.002 mol

3. Determine the ratio between FeSO4·7H2O and KMnO4 in the balanced equation:

From the balanced equation, we can see that 5 moles of FeSO4·7H2O react with 2 moles of KMnO4. Therefore, the ratio of moles of FeSO4·7H2O to moles of KMnO4 is 5:2.

4. Calculate the moles of FeSO4·7H2O in the sample:

To find the moles of FeSO4·7H2O in the sample, we can use the ratio from step 3. Multiply the moles of KMnO4 by the ratio of moles of FeSO4·7H2O to moles of KMnO4:

Moles of FeSO4·7H2O = 0.002 mol * (5/2) = 0.005 mol

5. Determine the molecular weight of FeSO4·7H2O:

The molecular weight of FeSO4·7H2O can be calculated by summing the atomic masses of its constituent elements:

Molecular weight of FeSO4·7H2O = (1 * atomic mass of Fe) + (1 * atomic mass of S) + (4 * atomic mass of O) + (14 * atomic mass of H) + (2 * atomic mass of H2O) = 278.02 g/mol

6. Calculate the mass of FeSO4·7H2O in the sample:

Mass of FeSO4·7H2O = Moles of FeSO4·7H2O * Molecular weight of FeSO4·7H2O

Mass of FeSO4·7H2O = 0.005 mol * 278.02 g/mol = 1.39 g

7. Calculate the percentage of FeSO4·7H2O in the sample:

Percentage of FeSO4·7H2O = (Mass of FeSO4·7H2O / Mass of sample) * 100

Given that the sample weighed 25 g, we can substitute these values into the equation:

Percentage of FeSO4·7H2O = (1.39 g / 25 g) * 100 = 5.56%

Therefore, the percentage of FeSO4·7H2O in the sample is found to be approximately 5.56%.