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An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:
- a)0.6
- b)0.4
- c)0.5
- d)0.8
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 2...
C2H6 + 3.5O2 → 2CO2 + 3H2O;
C2H4 + 3O2 → 2CO2 + 2H2O
Let volume of ethane is x litre,
22.4 × 4 = 3.5x + 3(28 – x)
⇒ x = 11.2 litre
at constant T and P, V αx n;
⇒ Mole fraction of C2H6 in mixture =
C2H4 + 3O2 → 2CO2 + 2H2O
Let volume of ethane is x litre,
22.4 × 4 = 3.5x + 3(28 – x)
⇒ x = 11.2 litre
at constant T and P, V αx n;
⇒ Mole fraction of C2H6 in mixture =
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An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 2...
Given:
Volume of mixture (V) = 28 L
Pressure (P) = 1 atm
Temperature (T) = 273 K
Mass of O2 (m) = 128 g
To find:
Mole fraction of C2H6 in the mixture
Solution:
1. Calculate the moles of O2 used in the reaction:
Molecular weight of O2 (M) = 2 × 16 = 32 g/mol
Number of moles of O2 (n) = m/M = 128/32 = 4 mol
2. Write the balanced chemical equation for the reaction:
C2H6 + 3O2 → 2CO2 + 3H2O
2C2H4 + 3O2 → 4CO2 + 4H2O
3. Calculate the moles of each reactant:
Since the mixture is ideal, we can use the ideal gas law to calculate the number of moles of each component:
PV = nRT
n = PV/RT
For ethane:
P = 1 atm
V = 28 L
R = 0.082 L atm/mol K
T = 273 K
n(C2H6) = (1 × 28)/(0.082 × 273) = 1.12 mol
For ethene:
n(C2H4) = 2 × n(C2H6) = 2.24 mol
4. Determine the limiting reactant:
From the balanced chemical equation, we can see that ethene requires 3/2 times the amount of O2 compared to ethane.
n(O2) = 4 mol
n(O2) required for ethane = (3/2) × n(C2H6) = (3/2) × 1.12 = 1.68 mol
n(O2) required for ethene = (3/2) × n(C2H4) = (3/2) × 2.24 = 3.36 mol
Since the amount of O2 available is less than the amount required for ethene, ethene is the limiting reactant.
5. Calculate the moles of CO2 and H2O produced:
From the balanced chemical equation, we can see that 2 moles of CO2 and 3 moles of H2O are produced for every mole of ethane reacted, and 4 moles of CO2 and 4 moles of H2O are produced for every mole of ethene reacted.
For ethene:
n(CO2) = 4 × n(C2H4) = 4 × 2.24 = 8.96 mol
n(H2O) = 4 × n(C2H4) = 4 × 2.24 = 8.96 mol
6. Calculate the mole fraction of C2H6:
Total moles of the mixture = n(C2H6) + n(C2H4) + n(O2) + n(CO2) + n(H2O)
= 1.12 + 2.24 + 4 + 8.96 + 8.96
= 25.28 mol
Mole fraction of C2H6
Volume of mixture (V) = 28 L
Pressure (P) = 1 atm
Temperature (T) = 273 K
Mass of O2 (m) = 128 g
To find:
Mole fraction of C2H6 in the mixture
Solution:
1. Calculate the moles of O2 used in the reaction:
Molecular weight of O2 (M) = 2 × 16 = 32 g/mol
Number of moles of O2 (n) = m/M = 128/32 = 4 mol
2. Write the balanced chemical equation for the reaction:
C2H6 + 3O2 → 2CO2 + 3H2O
2C2H4 + 3O2 → 4CO2 + 4H2O
3. Calculate the moles of each reactant:
Since the mixture is ideal, we can use the ideal gas law to calculate the number of moles of each component:
PV = nRT
n = PV/RT
For ethane:
P = 1 atm
V = 28 L
R = 0.082 L atm/mol K
T = 273 K
n(C2H6) = (1 × 28)/(0.082 × 273) = 1.12 mol
For ethene:
n(C2H4) = 2 × n(C2H6) = 2.24 mol
4. Determine the limiting reactant:
From the balanced chemical equation, we can see that ethene requires 3/2 times the amount of O2 compared to ethane.
n(O2) = 4 mol
n(O2) required for ethane = (3/2) × n(C2H6) = (3/2) × 1.12 = 1.68 mol
n(O2) required for ethene = (3/2) × n(C2H4) = (3/2) × 2.24 = 3.36 mol
Since the amount of O2 available is less than the amount required for ethene, ethene is the limiting reactant.
5. Calculate the moles of CO2 and H2O produced:
From the balanced chemical equation, we can see that 2 moles of CO2 and 3 moles of H2O are produced for every mole of ethane reacted, and 4 moles of CO2 and 4 moles of H2O are produced for every mole of ethene reacted.
For ethene:
n(CO2) = 4 × n(C2H4) = 4 × 2.24 = 8.96 mol
n(H2O) = 4 × n(C2H4) = 4 × 2.24 = 8.96 mol
6. Calculate the mole fraction of C2H6:
Total moles of the mixture = n(C2H6) + n(C2H4) + n(O2) + n(CO2) + n(H2O)
= 1.12 + 2.24 + 4 + 8.96 + 8.96
= 25.28 mol
Mole fraction of C2H6
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An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer?
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An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer?.
An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gaseous mixture of ethane (C2H6) and ethene (C2H4) occupies 28 litre at 1 atm and 273 K. The mixture reacts completely with 128 g O2 to produce CO2 and H2O. Mole fraction at C2H6 in the mixture is:a)0.6b)0.4c)0.5d)0.8Correct answer is option 'B'. Can you explain this answer?.
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