Question

(a) What volume of oxygen is required to burn completely a mixture of 22.4 dm³ of methane and 11.2 dm³ of hydrogen into carbon dioxide and steam? Equations of the reactions are given below:
CH₄ + 2O₂ $\stackrel{}{\to }$ CO₂ + 2H₂O
2H₂ + O₂ $\stackrel{}{\to }$ 2H₂O
(Assume that all volumes are measured at STP)

(b) The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular mass. Given 8 g of each gas at STP, which gas will contain the least number of molecules and which gas the most?

Answer 1

(a) 22.4 dm³ (or 22.4 L) of methane at STP means one mole of methane.
According to the given balanced equation:
One mole of methane requires two moles of oxygen for complete combustion.
Volume of oxygen required by methane = (2 $\times$ 22.4) dm³ = 44.8 dm³
Similarly, 11.2 dm³ (or 11.2 L) of hydrogen at STP means 0.5 moles of hydrogen.
0.5 moles of hydrogen require 0.25 moles of oxygen for complete combustion.
Volume of oxygen required by hydrogen = (0.25 $\times$ 22.4) dm³ = 5.6 dm³
Total volume of oxygen required = (5.6 + 44.8) dm³ = 50.4 dm³

(b) According to the ideal gas equation:
$$\begin{gathered} PV=n\mathrm{R}T \\ PV=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{gas}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{gas}}\times \mathrm{R}T \end{gathered}$$
At STP, higher the molecular mass of the gas, lower will be its volume. This in turn will decrease the number of molecules of the gas (according to the Avogadro's law).
Since chlorine has the highest molecular weight, it will have the lowest number of molecules. And because hydrogen has the smallest molecular weight, it will have the highest number of molecules.