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Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane is
- a)1.8 kg
- b)2.7 kg
- c)4.5 kg
- d)3.58 kg
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Amount of oxygen required for combustion of 1 kg of a mixture of butan...
Butane and isobutane have same molecular formula C4H10
C4H10 + 13/2 O2 → 4CO2 + 5H2O
58g C4H10 requires O2 = 13/2 x 32g
∴1000gC4H10 requires O2 = 13 x 32 x 1000/2 x 58
⇒ 358.62g
Or 3.586kg
Most Upvoted Answer
Amount of oxygen required for combustion of 1 kg of a mixture of butan...
**Combustion of Butane and Isobutane**
Butane (C4H10) and isobutane (C4H10) are both hydrocarbons that can undergo combustion reactions. In order for combustion to occur, a certain amount of oxygen is required to react with the hydrocarbon.
**Balanced Combustion Equation**
The balanced combustion equation for butane is:
2C4H10 + 13O2 → 8CO2 + 10H2O
And the balanced combustion equation for isobutane is:
C4H10 + 6.5O2 → 4CO2 + 5H2O
**Calculating the Amount of Oxygen**
To calculate the amount of oxygen required for the combustion of 1 kg of the mixture of butane and isobutane, we need to determine the proportion of each hydrocarbon in the mixture.
Let's assume the mixture contains x kg of butane and (1 - x) kg of isobutane.
The molecular weight of butane (C4H10) is 58 kg/kmol, and the molecular weight of isobutane (C4H10) is also 58 kg/kmol.
The number of moles of butane in the mixture is given by:
moles of butane = x kg / 58 kg/kmol = x/58 kmol
The number of moles of isobutane in the mixture is given by:
moles of isobutane = (1 - x) kg / 58 kg/kmol = (1 - x)/58 kmol
**Moles of Oxygen**
From the balanced combustion equations, we can see that each mole of butane requires 13 moles of oxygen, and each mole of isobutane requires 6.5 moles of oxygen.
Therefore, the moles of oxygen required for butane is:
moles of oxygen for butane = (x/58) kmol * 13 kmol O2/kmol C4H10 = 13x/58 kmol O2
And the moles of oxygen required for isobutane is:
moles of oxygen for isobutane = ((1 - x)/58) kmol * 6.5 kmol O2/kmol C4H10 = 6.5(1 - x)/58 kmol O2
**Total Moles of Oxygen**
The total moles of oxygen required for the mixture is the sum of the moles of oxygen required for butane and isobutane:
total moles of oxygen = moles of oxygen for butane + moles of oxygen for isobutane
total moles of oxygen = (13x/58) kmol O2 + 6.5(1 - x)/58 kmol O2
**Mass of Oxygen**
The mass of oxygen required can be calculated by multiplying the total moles of oxygen by the molecular weight of oxygen (32 kg/kmol):
mass of oxygen = total moles of oxygen * 32 kg/kmol
Substituting the expression for total moles of oxygen, we get:
mass of oxygen = [(13x/58) kmol O2 + 6.5(1 - x)/58 kmol O2] * 32 kg/kmol
Simplifying the expression:
mass of oxygen
Butane (C4H10) and isobutane (C4H10) are both hydrocarbons that can undergo combustion reactions. In order for combustion to occur, a certain amount of oxygen is required to react with the hydrocarbon.
**Balanced Combustion Equation**
The balanced combustion equation for butane is:
2C4H10 + 13O2 → 8CO2 + 10H2O
And the balanced combustion equation for isobutane is:
C4H10 + 6.5O2 → 4CO2 + 5H2O
**Calculating the Amount of Oxygen**
To calculate the amount of oxygen required for the combustion of 1 kg of the mixture of butane and isobutane, we need to determine the proportion of each hydrocarbon in the mixture.
Let's assume the mixture contains x kg of butane and (1 - x) kg of isobutane.
The molecular weight of butane (C4H10) is 58 kg/kmol, and the molecular weight of isobutane (C4H10) is also 58 kg/kmol.
The number of moles of butane in the mixture is given by:
moles of butane = x kg / 58 kg/kmol = x/58 kmol
The number of moles of isobutane in the mixture is given by:
moles of isobutane = (1 - x) kg / 58 kg/kmol = (1 - x)/58 kmol
**Moles of Oxygen**
From the balanced combustion equations, we can see that each mole of butane requires 13 moles of oxygen, and each mole of isobutane requires 6.5 moles of oxygen.
Therefore, the moles of oxygen required for butane is:
moles of oxygen for butane = (x/58) kmol * 13 kmol O2/kmol C4H10 = 13x/58 kmol O2
And the moles of oxygen required for isobutane is:
moles of oxygen for isobutane = ((1 - x)/58) kmol * 6.5 kmol O2/kmol C4H10 = 6.5(1 - x)/58 kmol O2
**Total Moles of Oxygen**
The total moles of oxygen required for the mixture is the sum of the moles of oxygen required for butane and isobutane:
total moles of oxygen = moles of oxygen for butane + moles of oxygen for isobutane
total moles of oxygen = (13x/58) kmol O2 + 6.5(1 - x)/58 kmol O2
**Mass of Oxygen**
The mass of oxygen required can be calculated by multiplying the total moles of oxygen by the molecular weight of oxygen (32 kg/kmol):
mass of oxygen = total moles of oxygen * 32 kg/kmol
Substituting the expression for total moles of oxygen, we get:
mass of oxygen = [(13x/58) kmol O2 + 6.5(1 - x)/58 kmol O2] * 32 kg/kmol
Simplifying the expression:
mass of oxygen
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Community Answer
Amount of oxygen required for combustion of 1 kg of a mixture of butan...
Butane and isobutane have same molecular formula C4H10C4H10 C4H10+132C4H10+132O2→4CO2+5H2OO2→4CO2+5H2O 58g C4H10C4H10 requires O2=132O2=132×32g×32g ∴1000gC4H10∴1000gC4H10 requires O2=13×32×10002×58O2=13×32×10002×58 ⇒358.62g⇒358.62g Or 3.586kg
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Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer?
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Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer?.
Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Amount of oxygen required for combustion of 1 kg of a mixture of butane and isobutane isa)1.8 kgb)2.7 kgc)4.5 kgd)3.58 kgCorrect answer is option 'D'. Can you explain this answer?.
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