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 = 12, O = 16, H = I, Molar Volume = 22.4 dm³ at stp.)

(i) Gram molecular mass of SO₂ = 32 + 2(16) = 32 + 32 = 64 g

64 g of SO₂ occupy 22.4 lit of vol.

320 g of SO₂ will occupy = $\dfrac{22.4}{64}$ x 320 = 112 lit.

Hence, volume of 320 g of SO₂ = 112 lit.

(ii) Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."

(iii)

$\begin{matrix} \text{C}$3\text{H}8 & + &5\text{O}2 & \longrightarrow & 3\text{CO}2 & + & 4\text{H}_2\text{O} \ 3(12) + 8(1) & & 5 \text{ vol} \ = 44 \text{ g}& & 5 (22.4) \text{ lit} \ \end{matrix}

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require $\dfrac{5 \times 22.4}{44}$ x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.