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Grade 10Physical Chemistry

(i) A sample of MnSO4. 4H2O is strongly heated in air. The residue is Mn3O4.
(ii) The residue is dissolved in 100 ml of 0.1 N FeSO4 containing dilute H2SO
(iii) The solution reacts completely with 50 ml of KMnO4 solution.
(iv) 25 ml of KMnO4 solution used in step (iii) requires 30 ml of 0.1 N FeSO4 solution for complete reaction.
Find the amount of MnSO4.4H2O present in the sample..

Profile image of Hrishant Goswami
12 Years agoGrade 10
Answers icon

1 Answer

Profile image of Jitender Pal
12 Years ago
Following reaction take place-
3MnSO4 4H2O
Mn2O4 + 4H2O ↑ + 3SO2 ↑ + O2 ↑
(residue)
Mn3O4 + 2FeSO4 + 4H2SO4 → Fe2(SO4)3 + 3MnSO4 + 4H2O
Milliequivalents of FeSO4 in 30 ml of 0.1N FeSO4 = 30 * 0.1 = 3 m.eq.
According to problem step (iv)
25 ml of KMnO4 reacts with = 3 m eq of FeSO4
Thus in step (iii) of the problems,
50 ml of KMnO4 reacts with = 3/25 * 50 m. eq. of FeSO4 = 6 meq of FeSO4
Milli eq. of 100 ml of 0.1N FeSO4 = 100 * 0.1 = 10 meq.
FeSO4 wich reacted with Mn3O4 = (10-6) = 4m eq.
Milli eq of FeSO4 = Millie q. of Mn3O4
(∵ Millie q of oxidizing agent and reducing agent are equal)
∵ Mn3O4 ≡ 3Mn3SO4.4H2O
∴ 1 Meq of Mn3O4 = 3 Meq of MnSO4 4H2O
∴ 4 Meq of Mn3O4 = 12 Meq of MnSO4 4H2O
Eq. wt of MnSO4 4H2O = Mol wt./2 = 223/2 = 111.5
Wt. of MnSO4 4H2O in sample = 12 * 111.5 = 1338 mg = 1.338g.

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