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The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.
- a)220
- b)340
- c)440
- d)560
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is ...
There is one Carbon atom in the molecule of CO2.
Volume of C4 H10 in the mixture = 80 ml
volume of CO2 produced by combustion of C4H10 = 4 * 80 = 320 ml
as there are 4 Carbons in each molecule.
volume of CH4 + CO in mixture = 120 ml
volume of CO2 produced = 120 ml
Total volume of CO2 = 320 + 120 = 440 ml
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The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is ...
Given: % by volume of C4H10 in the mixture = 40
To find: Volume of CO2 produced when 200 ml of the mixture is burnt in excess of O2
Let's assume the total volume of the mixture as V ml.
Volume % of C4H10 in the mixture = 40
Volume % of CH4 in the mixture = (100 - 40)/2 = 30
Volume % of CO in the mixture = (100 - 40 - 30) = 30
So, the volume of C4H10 in the mixture = (40/100) x V ml = 0.4V ml
The volume of CH4 in the mixture = (30/100) x V ml = 0.3V ml
The volume of CO in the mixture = (30/100) x V ml = 0.3V ml
When the mixture is burnt in excess of O2, the following reaction takes place:
C4H10 + 13/2 O2 → 4CO2 + 5H2O
From the balanced chemical equation, we can see that for every 1 volume of C4H10, 4 volumes of CO2 are produced.
So, the volume of CO2 produced from C4H10 = 4 x 0.4V ml = 1.6V ml
Similarly, the volume of CO2 produced from CH4 and CO can be calculated as follows:
CH4 + 2O2 → CO2 + 2H2O
For every 1 volume of CH4, 1 volume of CO2 is produced.
So, the volume of CO2 produced from CH4 = 1 x 0.3V ml = 0.3V ml
CO + 1/2 O2 → CO2
For every 1 volume of CO, 1 volume of CO2 is produced.
So, the volume of CO2 produced from CO = 1 x 0.3V ml = 0.3V ml
Therefore, the total volume of CO2 produced when 200 ml of the mixture is burnt in excess of O2 is:
Total volume of CO2 = Volume of CO2 produced from C4H10 + Volume of CO2 produced from CH4 + Volume of CO2 produced from CO
Total volume of CO2 = 1.6V ml + 0.3V ml + 0.3V ml
Total volume of CO2 = 2.2V ml
As we know, the total volume of the mixture is V = 200 ml, we can substitute V = 200 ml in the above equation to get the total volume of CO2 produced.
Total volume of CO2 = 2.2 x 200 ml = 440 ml
Hence, the correct option is (c) 440 ml.
To find: Volume of CO2 produced when 200 ml of the mixture is burnt in excess of O2
Let's assume the total volume of the mixture as V ml.
Volume % of C4H10 in the mixture = 40
Volume % of CH4 in the mixture = (100 - 40)/2 = 30
Volume % of CO in the mixture = (100 - 40 - 30) = 30
So, the volume of C4H10 in the mixture = (40/100) x V ml = 0.4V ml
The volume of CH4 in the mixture = (30/100) x V ml = 0.3V ml
The volume of CO in the mixture = (30/100) x V ml = 0.3V ml
When the mixture is burnt in excess of O2, the following reaction takes place:
C4H10 + 13/2 O2 → 4CO2 + 5H2O
From the balanced chemical equation, we can see that for every 1 volume of C4H10, 4 volumes of CO2 are produced.
So, the volume of CO2 produced from C4H10 = 4 x 0.4V ml = 1.6V ml
Similarly, the volume of CO2 produced from CH4 and CO can be calculated as follows:
CH4 + 2O2 → CO2 + 2H2O
For every 1 volume of CH4, 1 volume of CO2 is produced.
So, the volume of CO2 produced from CH4 = 1 x 0.3V ml = 0.3V ml
CO + 1/2 O2 → CO2
For every 1 volume of CO, 1 volume of CO2 is produced.
So, the volume of CO2 produced from CO = 1 x 0.3V ml = 0.3V ml
Therefore, the total volume of CO2 produced when 200 ml of the mixture is burnt in excess of O2 is:
Total volume of CO2 = Volume of CO2 produced from C4H10 + Volume of CO2 produced from CH4 + Volume of CO2 produced from CO
Total volume of CO2 = 1.6V ml + 0.3V ml + 0.3V ml
Total volume of CO2 = 2.2V ml
As we know, the total volume of the mixture is V = 200 ml, we can substitute V = 200 ml in the above equation to get the total volume of CO2 produced.
Total volume of CO2 = 2.2 x 200 ml = 440 ml
Hence, the correct option is (c) 440 ml.
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The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.a)220b)340c)440d)560Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.a)220b)340c)440d)560Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The % by volume of C4H10 in a gaseous mixture of C4H10, CH4 and CO is 40. When 200 ml of the mixture is burnt in excess of O2. Find volume (in ml) of CO2 produced.a)220b)340c)440d)560Correct answer is option 'C'. Can you explain this answer?.
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