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Question

# (i) Calculate the volume of 320 g of SO2 at stp (Atomic mass: S = 32 and O = 16) (ii) State Gay-Lussac's Law of combining volumes. (iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8). (Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm3 at stp)

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Solution

## (i) Number of moles of SO2 in 320 g of SO2 = $\frac{\mathrm{Mass}\mathrm{of}{\mathrm{SO}}_{2}}{\mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{SO}}_{2}}=\frac{320}{\left(32+\left(16×2\right)\right)}=5\mathrm{mol}$ At STP, one mole of SO2 occupies 22.4 L of volume. ∴ Volume of five moles of SO2 = (5 $×$ 22.4) L = 112 L (ii) Gay-Lussac's law states that reacting volumes of the gases bear a simple ratio with each other and to the volume of the gaseous products, when volumes are measured at same temperature and pressure. (iii) Reaction of combustion of propane is: C3H8 + 5O2 $\to$ 3CO2 + 4H2O So, for combustion of one mole of propane, five moles of oxygen is required. Number of moles of propane in 8.8 g = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{propane}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{propane}}=\frac{8.8}{\left(\left(3×12\right)+\left(8×1\right)\right)}=0.2\mathrm{mol}$ Number of moles of oxygen consumed by 0.2 mole of propane = (5 $×$ 0.2) mol = 1 mol One mole of oxygen at STP will occupy 22.4 L or 22.4 dm3.

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