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### AuthorTopic: Stoichiometry question: combustion of natural gas  (Read 223 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### supernatant

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##### Stoichiometry question: combustion of natural gas
« on: January 12, 2018, 05:22:58 AM »

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?
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#### mjc123

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##### Re: Stoichiometry question: combustion of natural gas
« Reply #1 on: January 12, 2018, 07:07:07 AM »

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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#### supernatant

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##### Re: Stoichiometry question: combustion of natural gas
« Reply #2 on: January 12, 2018, 07:36:32 AM »

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.
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