Question

(i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:
C₃H_g(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)
2C₄H_10(g) + 13O_2(g) → 8CO_2(g) + 10H₂O_(g)

(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16].

Answer 1

(i) Volume of LPG = 10 L
Volume of propane in 10 litres of LPG = $\frac{60}{100}\times 10=6\mathrm{L}$
Volume of butane in 10 litres of LPG = $\frac{40}{100}\times 10=4\mathrm{L}$
During combustion, one mole of propane gives three moles of carbon dioxide.
Number of moles of carbon dioxide liberated from six litres of propane can be calculated from the ideal gas equation:
PV = nRT
Number of moles of propane in six litres of propane = $\frac{P\times 6}{\mathrm{R}T}$
Number of moles of carbon dioxide liberated = $3\times \left(\frac{P\times 6}{\mathrm{R}T}\right)$
Volume of carbon dioxide = $\frac{n\mathrm{R}T}{P}$
Volume of carbon dioxide = $3\times \left(\frac{P\times 6}{\mathrm{R}T}\right)\times \left(\frac{\mathrm{R}T}{P}\right)$ = 18 L
Number of moles of carbon dioxide liberated from one mole of butane = 4
Volume of carbon dioxide liberated from butane is four times the volume of butane = 16 L
Total volume of carbon dioxide liberated = 18 + 16 = 34 L

(ii) Molecular mass of ammonium nitrate (NH₄)NO₃ = 80 g
Mass of nitrogen in ammonium nitrate = (14 + 14) g = 28 g
$\mathrm{Percentage}\mathrm{of}\mathrm{N}\mathrm{in}{\mathrm{NH}}_4{\mathrm{NO}}_3=\frac{\mathrm{Mass}\mathrm{of}\mathrm{N}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_4{\mathrm{NO}}_3}\times 100=\frac{28}{80}\times 100=35\%$

Mass of oxygen in ammonium nitrate = 3(16) g = 48 g
$\mathrm{Percentage}\mathrm{of}\mathrm{O}\mathrm{in}{\mathrm{NH}}_4{\mathrm{NO}}_3=\frac{\mathrm{Mass}\mathrm{of}\mathrm{O}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_4{\mathrm{NO}}_3}\times 100=\frac{48}{80}\times 100=60\%$