Question

5.5g of a mixture of feSO4.7h2O and fe(SO4)3.9H2O required 5.4 ml of 0.1 N KMnO4 solution for complete oxidation.calculate mole of hydrated ferric sulphate in mixture.

Answer 1

Let mass of ferrous sulphate is x g so mass of ferric sulphate is 5.5- x g Since only ferrous ion oxidized therefore me of ferrous sulphate = me of KMnO4 Ie x/ 139 x 1000= 5.4 x 0.1 On solving we get x= 0.07506 Therefore moles of ferric salt = 5.5-0.07506/ 562 = 9.65x 10^-3

Answer 2

Answer:9.5 x 10^-3

Explanation:let FeSO4.7H2O is X gm then Fe2(SO4)3.9H2O is (5.5-X)gm.so eq. of FeSO4.7H2O=eq. of  Fe2(SO4)3.9H2O

X/278 *1 =5.4/1000 *0.1

X=0.15 gm

weight of FeSO4.7H2O= 0.15 gm

so  weight of Fe2(SO4)3.9H2O=5.5-0.15= 5.35gm

moles of Fe2(SO4)3.9H2O= 5.35/562 =9.5 x 10^-3