Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: supernatant on January 12, 2018, 11:22:58 AM
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Hi all! Here's a question I've been struggling with for a few hours:
A typical composition of natural gas could be 90% methane, 8% ethane and 2% propane (mol). Under standard conditions for temperature and pressure, the complete combustion of 11.2 m3 of natural gas produces, approximately, the following mass of CO2:
a) 22.4 kg
b) 11.2 kg
c) 32.6 kg
d) 24.6 kg
Letter d is the correct answer. This is what I did:
1 mol of any gas occupies the volume of 22.4 L. Therefore, 11.2 m3 is equal to 500 mol of natural gas. So I wrote the chemical equation:
1 CH4 + 1 C2H6 + 1 C3H8 + 21/2 O2 :rarrow: 6 CO2 + 9 H2O
Considering the proportions, 1 mol of natural gas produces 6 mol of CO2. So 500 mol produces 3000 mol of CO2, which is equivalent to 132 kg.
What am I doing wrong here?
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You have 3 moles of hydrocarbons on the LHS (1+1+1); 1 mole of this mixture produces 2 moles of CO2
But why do you write that equation? You are told the composition of natural gas, and it is not 1:1:1. Try using the correct proportions. (As a rough estimate, since it is 90% methane, see what you get assuming it is all methane.)
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I got it now, thank you! :)
Due to the different proportions between the alkanes and carbon dioxide, I had to consider the equations separately:
1 CH4 + 2 O2 :rarrow: 1 CO2 + 2 H2O
90% of 500 mol = 450 mol
450 mol of CH4 produces 450 mol of CO2
1 C2H6 + 7/2 O2 :rarrow: 2 CO2 + 3 H2O
8% of 500 mol = 40 mol
40 mol of CH4 produces 80 mol of CO2
1 C3H8 + 5 O2 :rarrow: 3 CO2 + 4 H2O
2% of 500 mol = 10 mol
10 mol of CH4 produces 30 mol of CO2
450 + 80 + 30 = 560 moles of CO2 are produced.
560 mol x 44 g/mol = 24640 g = 24,6 kg
If you assume it's all methane, you get 22,0 kg of carbon dioxide.