Air contains 79% N2 and 21% O2 on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2 in the products is::
Question
Air contains 79% N2 and 21% O2 on a molar basis. Methane (CH4) is burned with 50% excess air than required stoichiometrically. Assuming complete combustion of methane, the molar percentage of N2 in the products is::
Solution
The stoichiometric combustion reaction of methane with air is:
CH4 + 2(O2 + 3.76N2) -> CO2 + 2H2O + 2*3.76N2
This means that for each mole of methane, we need 2 moles of oxygen for complete combustion. However, the problem states that there is 50% excess air, so instead of 2 moles of oxygen, we have 3 moles of oxygen (2 moles + 50% of 2 moles).
The nitrogen in the air is not involved in the combustion reaction, so it remains as a product. For each mole of methane, we have 23.76 moles of nitrogen in the stoichiometric case, and 33.76 moles of nitrogen in the 50% excess air case.
The total moles of products in the 50% excess air case is:
1 mole of CO2 (from the combustion of methane) 2 moles of H2O (from the combustion of methane) 3*3.76 moles of N2 (from the excess air)
Adding these up, we get a total of 1 + 2 + 3*3.76 = 13.28 moles of products.
The molar percentage of N2 in the products is therefore (3*3.76/13.28)*100 = 85%.
So, the molar percentage of N2 in the products is 85%.
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