Question

(a) 15.8 g of potassium permanganate (VII) was used in the reaction given below:
2KMnO, + 10 FeSO4 + 8H2SO4 →K2SO4, + 2MnSO4 + 5Fe2 (SO4)3 +8H20
Calculate the mass of Iron (II) sulphate used in the above reaction.
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Answer 1

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ANSWER

Molecular weight of KMnO

4

=39+55+164=158

Molecular weight of K

2

SO

4

=239+32+164=174

Molecular weight of FeSO

4

=56+32+164=152

2×158 g of KMnO

4

yields 174 g of K

2

SO

4

15.8 g of KMnO

4

will yield [(174)/(2×158)]×15.8 g = 8.7 g of K

2

SO

4

.

174 g of K

2

SO

4

yields 1520 g of FeSO

4

8.7 g of K

2

SO

4

will yield (1520/174)×8.7 = 76 g of FeSO

4

.

Hence, 76 g of Iron (II) sulphate is used

Answer 2

Answer:

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