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प्रश्न
The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMno4 + 10FeSO4 + 8H2O → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O
If 15.8g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.
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उत्तर
Molecular weight of KMnO4 = 39 + 55 + 16 x 4 = 158
Molecular weight of K2SO4 = 2 x 39 + 32 + 16 x 4 = 174
Molecular weight of FeSO4 = 56 + 32 + 64 = 152
2 x 158g of KMnO4 yields= 174g of K2SO4.
So, 15.8g of KMnO4 will yield = 174 x 15.8/2 x 158 = 8.7g of K2SO4 .
174g of K2SO4 yields 152g of FeSO4
So, 8.7 g of K2SO4 will yield = 152 x 8.7/174 = 7.6 g of FeSO4.
Hence, 7.6g of iron (II) sulphate is used in the above reaction.
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