[supernatant]

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  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?

[mjc123]

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.)

[supernatant]

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 1 CO₂ + 2 H₂O

90% of 500 mol = 450 mol
450 mol of CH₄ produces 450 mol of CO₂

1 C₂H₆ + 7/2 O₂ 2 CO₂ + 3 H₂O

8% of 500 mol = 40 mol
40 mol of CH₄ produces 80 mol of CO₂

1 C₃H₈ + 5 O₂ 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.