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Difficulty: HardCarbon: Allotropes, Coal Distillation, and Industrial Fuel Gases

During the destructive distillation of coal, the volatile gaseous fraction is collected and purified to yield coal gas. A 50.0 dm350.0\text{ dm}^3 sample of this coal gas contains 50.0%50.0\% hydrogen (H2\text{H}_2), 30.0%30.0\% methane (CH4\text{CH}_4), and 20.0%20.0\% carbon(II) oxide (CO\text{CO}) by volume. What is the total volume of pure oxygen gas required at STP for the complete combustion of this sample?

  1. 47.5 dm347.5\text{ dm}^3Answer
  2. B
    32.5 dm332.5\text{ dm}^3
  3. C
    60.0 dm360.0\text{ dm}^3
  4. D
    95.0 dm395.0\text{ dm}^3

Answer

The total volume of pure oxygen required at STP for complete combustion is 47.5 dm347.5\text{ dm}^3.
By Gay-Lussac's law of combining volumes, gases react in simple whole-number volume ratios. In a 50.0 dm350.0\text{ dm}^3 coal gas mixture, the volumes of H2\text{H}_2, CH4\text{CH}_4, and CO\text{CO} are 25.0 dm325.0\text{ dm}^3, 15.0 dm315.0\text{ dm}^3, and 10.0 dm310.0\text{ dm}^3 respectively. According to their balanced combustion equations, H2\text{H}_2 requires half its volume in O2\text{O}_2 (12.5 dm312.5\text{ dm}^3), CH4\text{CH}_4 requires twice its volume in O2\text{O}_2 (30.0 dm330.0\text{ dm}^3), and CO\text{CO} requires half its volume in O2\text{O}_2 (5.0 dm35.0\text{ dm}^3). Summing these gives 12.5+30.0+5.0=47.5 dm312.5 + 30.0 + 5.0 = 47.5\text{ dm}^3.

Step-by-Step Solution

1
Calculate the individual volumes of each constituent gas in the 50.0 dm350.0\text{ dm}^3 coal gas mixture.
Volume of H2=0.500×50.0 dm3=25.0 dm3\text{H}_2 = 0.500 \times 50.0\text{ dm}^3 = 25.0\text{ dm}^3; Volume of CH4=0.300×50.0 dm3=15.0 dm3\text{CH}_4 = 0.300 \times 50.0\text{ dm}^3 = 15.0\text{ dm}^3; Volume of CO=0.200×50.0 dm3=10.0 dm3\text{CO} = 0.200 \times 50.0\text{ dm}^3 = 10.0\text{ dm}^3.
By Gay-Lussac's Law, volume percentages directly represent mole fractions at constant temperature and pressure.
2
Write the balanced chemical equations for the complete combustion of each component.
(1) 2H2(g)+O2(g)→2H2O(l)2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}
(2) \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}(3) (3) 2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}$.
Stoichiometric coefficients define the combining volume ratios of reactants.
3
Determine the volume of O2\text{O}_2 needed for each gas component.
For H2\text{H}_2: V(O2)=12×25.0 dm3=12.5 dm3V(\text{O}_2) = \frac{1}{2} \times 25.0\text{ dm}^3 = 12.5\text{ dm}^3.
For CH4\text{CH}_4: V(O2)=2×15.0 dm3=30.0 dm3V(\text{O}_2) = 2 \times 15.0\text{ dm}^3 = 30.0\text{ dm}^3.
For CO\text{CO}: V(O2)=12×10.0 dm3=5.0 dm3V(\text{O}_2) = \frac{1}{2} \times 10.0\text{ dm}^3 = 5.0\text{ dm}^3.
Applying Gay-Lussac's law of combining volumes to each combustion reaction.
4
Sum the required volumes of oxygen for all three components.
Total V(O2)=12.5 dm3+30.0 dm3+5.0 dm3=47.5 dm3V(\text{O}_2) = 12.5\text{ dm}^3 + 30.0\text{ dm}^3 + 5.0\text{ dm}^3 = 47.5\text{ dm}^3.
The total oxygen needed is the additive sum of individual combustion demands.

Key Concept

Combustion Stoichiometry of Industrial Fuel Gases
Estimated Time:2m 0s
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Topics

Carbon: Allotropes, Coal Distillation, and Industrial Fuel Gases

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