This is a two part question to which I\'ve answered the first part but am unsure

Question

This is a two part question to which I've answered the first part but am unsure about how to proceed with the second part.

Part 1: Write a balanced combustion reaction for natural gas composed of 90 mol% methane and 10 mol% ethane. [Fractional moles are convenient for the subsequents calculations. Your reaction should have .90 mol methane and .10 mol ethane on the left side. Do not include nitrogen.]

The answer I've obtained for this part of the question is:

0.9CH4 + 0.2 C2H6 + 2.5 O2 --> 2.4 H2O + 1.3 CO2

Part 2: Now note that air (not pure oxygen) flows through the combustion chamber. Rewrite your combustion reaction from Part 1 to include all molecules present, whether or not they participate in the reaction. Assume air is 80 mol% Nitrogen (N2) and 20 mol% Oxygen (O2). Assume no excess air is present.

Explanation / Answer

80 % mol of N2 and 20 % mol O2 means 1 part of O2 per 4 parts N2

and hence eq becomes

0.9 CH4 + 0.2 C2H6 + 2.5 O2 + 10 N2 ------------> 2.4 H2O + 1.3 CO2 + 10 N2

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