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Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?
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Chemistry question A 44.00g sample of a natural gas, consisting of met...
Problem:
A 44.00g sample of natural gas consisting of methane (CH4) and ethylene (C2H4) was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?
Solution:
To determine the percent of ethylene in the sample, we need to follow the steps below:
Step 1: Write and balance the chemical equation for the combustion of natural gas.
CH4 + 2C2H4 + 3O2 → 4CO2 + 2H2O
Step 2: Calculate the moles of CO2 produced.
132g CO2 x (1 mol CO2 / 44.01 g CO2) = 3.00 mol CO2
Step 3: Determine the limiting reagent.
To determine the limiting reagent, we need to calculate the moles of CH4 and C2H4 in the sample.
Moles of CH4 = 44.00 g / 16.04 g/mol = 2.744 mol
Moles of C2H4 = (44.00 g - 2.744 mol x 28.05 g/mol) / 42.08 g/mol = 0.490 mol
The limiting reagent is CH4 because it produces less CO2 compared to C2H4. The stoichiometric ratio for CH4 to CO2 is 1:1, while the stoichiometric ratio for C2H4 to CO2 is 2:4 or 1:2.
Step 4: Calculate the moles of C2H4 in the sample.
Since CH4 is the limiting reagent, the moles of CO2 produced come from the combustion of CH4 only. Therefore, the moles of C2H4 in the sample are:
Moles of C2H4 = (3.00 mol CO2 / 1 mol CH4) x (2 mol C2H4 / 4 mol CO2) = 0.750 mol
Step 5: Calculate the percent of ethylene in the sample.
The percent of ethylene in the sample is:
Percent of C2H4 = (0.490 mol + 0.750 mol) / 44.00 g x 100% = 3.09%
Therefore, the percent of ethylene in the sample is 3.09%.
Conclusion:
The percent of ethylene in the natural gas sample is 3.09%.
A 44.00g sample of natural gas consisting of methane (CH4) and ethylene (C2H4) was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?
Solution:
To determine the percent of ethylene in the sample, we need to follow the steps below:
Step 1: Write and balance the chemical equation for the combustion of natural gas.
CH4 + 2C2H4 + 3O2 → 4CO2 + 2H2O
Step 2: Calculate the moles of CO2 produced.
132g CO2 x (1 mol CO2 / 44.01 g CO2) = 3.00 mol CO2
Step 3: Determine the limiting reagent.
To determine the limiting reagent, we need to calculate the moles of CH4 and C2H4 in the sample.
Moles of CH4 = 44.00 g / 16.04 g/mol = 2.744 mol
Moles of C2H4 = (44.00 g - 2.744 mol x 28.05 g/mol) / 42.08 g/mol = 0.490 mol
The limiting reagent is CH4 because it produces less CO2 compared to C2H4. The stoichiometric ratio for CH4 to CO2 is 1:1, while the stoichiometric ratio for C2H4 to CO2 is 2:4 or 1:2.
Step 4: Calculate the moles of C2H4 in the sample.
Since CH4 is the limiting reagent, the moles of CO2 produced come from the combustion of CH4 only. Therefore, the moles of C2H4 in the sample are:
Moles of C2H4 = (3.00 mol CO2 / 1 mol CH4) x (2 mol C2H4 / 4 mol CO2) = 0.750 mol
Step 5: Calculate the percent of ethylene in the sample.
The percent of ethylene in the sample is:
Percent of C2H4 = (0.490 mol + 0.750 mol) / 44.00 g x 100% = 3.09%
Therefore, the percent of ethylene in the sample is 3.09%.
Conclusion:
The percent of ethylene in the natural gas sample is 3.09%.
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Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?
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Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?.
Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Chemistry question A 44.00g sample of a natural gas, consisting of methane (CH4) and ethylene (C2H4), was burned in excess oxygen, yielding 132g CO2 and some H2O as products. What percent of the sample is ethylene?.
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