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Question

# (a) A flask contains 3.2 g of sulphur dioxide. Calculate the following: (i) The moles of sulphur dioxide present in the flask. (ii) The number of molecules of sulphur dioxide present in the flask. (iii) The volume occupied by 3.2 g of sulphur dioxide at STP. (S = 32, O = 16) (b) The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below: 2KMnO4 + 10FeSO4 + 8H2SO4 $\stackrel{}{\to }$ K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O If 15.8 g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction. (K = 39, Mn = 55, Fe = 56, S = 32, O = 16)

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Solution

## (a) (i) Moles of sulphur dixoide in 3.2 g = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{sulphur}\mathrm{dioxide}}{\mathrm{Molecular}\mathrm{weight}\mathrm{of}\mathrm{sulphur}\mathrm{dioxide}}=\frac{3.2}{64}=0.05$ mol (ii) Number of molecules in 0.05 moles of sulphur dioxide = Number of moles $×$ Avogadro's number = 0.05 $×$ 6.023 $×$ 1023 = 3.01 $×$1022 (iii) At STP, one mole of sulphur dioxide occupies 22.4 L of volume. Volume of 0.05 mole of sulphur dioxide at STP = (22.4 $×$ 0.05) L = 1.12 L (b) According to the given balanced equation, for two moles of potassium permanganate, 10 moles of iron (II) sulphate are used. Alternatively, for one mole of potassium permanganate, five moles of iron (II) sulphate are used. Number of moles of potassium permanganate in 15.8 g = $\frac{15.8}{39+55+4\left(16\right)}=\frac{15.8}{158}=0.1$ (Number of moles of iron (II) sulphate used by 0.1 moles of potassium permanganate = (5 $×$ 0.1) = 0.5 mol Mass of iron (II) sulphate in 0.5 moles = [0.5 $×$ (56 + 32 + 64)] g = (0.5 $×$ 152) g = 76 g

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