Home
Class 12
CHEMISTRY
5.5 g of a mixutre of FeSO4.7H2O and Fe2...

5.5 g of a mixutre of `FeSO_4.7H_2O` and `Fe_2(SO_4)_3.9H_2O` requires 5.4 " mL of " `0.1 N KMnO_4` solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture.

Text Solution

Verified by Experts

Equivalent of `KMnO_(4)` = Equivalents of `FeSO_(4).7H_(2)O`
5.4 ml of 0.1 N `KMnO_(4) = 5.4 xx 10^(-4)`equivalents
Amount of `FeSO_(4) = 5.4 xx 10^(-4) xx MoLwt. of FeSO_(4).7H_(2)O = 5.4 xx 10^(-4) xx 278 = 0.150 g`
Total weight of mixture = 5.5 g
Amount of ferric sulphate = 5.5 – 0.150 g = 5.35 g
Hence moles of ferric sulphate = `Mass/MoLwt. = 5.35/562 = 9.5 xx 10^(-3)` gram-mole.
Promotional Banner

Similar Questions

Explore conceptually related problems

3 " mol of "a mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) requried 100 " mL of " 2 M KMnO_(4) solution in acidic medium. Hence, mole fraction of FeSO_(4) in the mixture is

A 3 mole mixture of FeSO_(4) and Fe_(2)(SO_(4))_(3) required 100 mL of 2M KMnO_(4) solution in acidic medium. Find the mole of FeSO_(4) in the mixture.

1 mole of equimolar mixture of Fe_(2)(C_(2)O_(4))_(3) and FeC_(2)O_(4) required X moles of KMnO_(4) in acid medium for complete reaction. The value of X is:

A mixture of H_(2)C_(2)O_(4) and K_(2)C_(2)O_(4) required 0.2 N, 25mL KMnO_(4) solution for complete oxidation. Same mixture needs 0.2 M,20 mL NaOH solution for its complete neutralisation. Calculate mole percentage of H_(2)C_(2)O_(4) in the given mixture.