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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 ________________
Correct answer is between '73,74'. Can you explain this answer?
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Air contains 79% N2 and 21% O2 on a molar basis. Methane (CH4) is burn...
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Air contains 79% N2 and 21% O2 on a molar basis. Methane (CH4) is burn...
Given:
- Air contains 79% N2 and 21% O2 on a molar basis.
- Methane (CH4) is burned with 50% excess air than required stoichiometrically.
- Complete combustion of methane.
To find:
The molar percentage of N2 in the products.
Solution:
1. Determine the stoichiometric ratio of methane to air:
- The stoichiometric ratio of methane to air can be calculated using the balanced chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
- From the balanced equation, we can see that 1 mole of CH4 reacts with 2 moles of O2.
- Therefore, the stoichiometric ratio of methane to air is 1:2.
2. Calculate the amount of air required stoichiometrically:
- Since the stoichiometric ratio of methane to air is 1:2, the amount of air required stoichiometrically is half the amount of methane.
- Let's assume we have 1 mole of methane, then the amount of air required stoichiometrically would be 2 moles.
3. Calculate the amount of excess air:
- The excess air is given as 50% more than required stoichiometrically.
- 50% excess air means an additional 50% of 2 moles of air required stoichiometrically.
- Therefore, the amount of excess air is 1.5 * 2 = 3 moles of air.
4. Determine the total moles of air in the products:
- Since the excess air is burned along with methane, the total moles of air in the products would be the sum of the air required stoichiometrically and the excess air.
- Total moles of air = 2 moles (stoichiometric air) + 3 moles (excess air) = 5 moles.
5. Calculate the molar percentage of N2 in the products:
- The molar percentage of N2 can be calculated by dividing the moles of N2 by the total moles of air and multiplying by 100.
- Moles of N2 = 79% of 5 moles = 3.95 moles.
- Molar percentage of N2 = (3.95 moles / 5 moles) * 100 = 79%.
Answer:
The molar percentage of N2 in the products is 79%.
- Air contains 79% N2 and 21% O2 on a molar basis.
- Methane (CH4) is burned with 50% excess air than required stoichiometrically.
- Complete combustion of methane.
To find:
The molar percentage of N2 in the products.
Solution:
1. Determine the stoichiometric ratio of methane to air:
- The stoichiometric ratio of methane to air can be calculated using the balanced chemical equation for the combustion of methane:
CH4 + 2O2 -> CO2 + 2H2O
- From the balanced equation, we can see that 1 mole of CH4 reacts with 2 moles of O2.
- Therefore, the stoichiometric ratio of methane to air is 1:2.
2. Calculate the amount of air required stoichiometrically:
- Since the stoichiometric ratio of methane to air is 1:2, the amount of air required stoichiometrically is half the amount of methane.
- Let's assume we have 1 mole of methane, then the amount of air required stoichiometrically would be 2 moles.
3. Calculate the amount of excess air:
- The excess air is given as 50% more than required stoichiometrically.
- 50% excess air means an additional 50% of 2 moles of air required stoichiometrically.
- Therefore, the amount of excess air is 1.5 * 2 = 3 moles of air.
4. Determine the total moles of air in the products:
- Since the excess air is burned along with methane, the total moles of air in the products would be the sum of the air required stoichiometrically and the excess air.
- Total moles of air = 2 moles (stoichiometric air) + 3 moles (excess air) = 5 moles.
5. Calculate the molar percentage of N2 in the products:
- The molar percentage of N2 can be calculated by dividing the moles of N2 by the total moles of air and multiplying by 100.
- Moles of N2 = 79% of 5 moles = 3.95 moles.
- Molar percentage of N2 = (3.95 moles / 5 moles) * 100 = 79%.
Answer:
The molar percentage of N2 in the products is 79%.
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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 ________________Correct answer is between '73,74'. Can you explain this answer?
Question Description
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 ________________Correct answer is between '73,74'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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 ________________Correct answer is between '73,74'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 ________________Correct answer is between '73,74'. Can you explain this answer?.
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 ________________Correct answer is between '73,74'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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 ________________Correct answer is between '73,74'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 ________________Correct answer is between '73,74'. Can you explain this answer?.
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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 ________________Correct answer is between '73,74'. Can you explain this answer?, a detailed solution for 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 ________________Correct answer is between '73,74'. Can you explain this answer? has been provided alongside types of 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 ________________Correct answer is between '73,74'. Can you explain this answer? theory, EduRev gives you an
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