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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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Copy the following table which gives the volumes of the gases collected, and the number of molecules (X) in 20L of nitrogen.You are to complete the table by giving the number of molecules in th e other gases, in terms of X.
| Gas | Volume(litres) | Number of molecules |
| Chlorine | 10 | |
| Nitrogen | 20 | X |
| Ammonia | 20 | |
| Sulphur dioxide | 5 |
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When heated, potassium permanganate decomposes according to the following equation :
\[\ce{2KMnO4 -> \underset{\text{solid residue}}{K2MnO4 + MnO2} + O2}\]
(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
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