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)

Answer 1

(i) Number of moles of SO₂ in 320 g of SO₂ = $\frac{\mathrm{Mass}\mathrm{of}{\mathrm{SO}}_2}{\mathrm{Molecular}\mathrm{weight}\mathrm{of}{\mathrm{SO}}_2}=\frac{320}{\left(32+\left(16\times 2\right)\right)}=5\mathrm{mol}$
At STP, one mole of SO₂ occupies 22.4 L of volume.
∴ Volume of five moles of SO₂ = (5 $\times$ 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 = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{propane}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{propane}}=\frac{8.8}{\left(\left(3\times 12\right)+\left(8\times 1\right)\right)}=0.2\mathrm{mol}$
Number of moles of oxygen consumed by 0.2 mole of propane = (5 $\times$ 0.2) mol = 1 mol
One mole of oxygen at STP will occupy 22.4 L or 22.4 dm³.