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200 ml of a gaseous mixture containing CO, CO2 and N2 on complete combustion in just sufficient amount of O2 showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2 :VCO : VN2 in original mixture.
- a)4 : 1 : 5
- b)2 : 3 : 5
- c)1 : 4 : 5
- d)1 : 3 : 5
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
200 ml of a gaseous mixture containing CO, CO2and N2on complete combus...
Volume of mixture of CO, CO2 and N2 = 200 ml
Volume of CO = x ml
Volume of N2 = y ml
Volume of CO2 = 200 - x - y ml
On combustion, CO2 remains as it is. Nitrogen burns only at very
high temperatures. At low temperatures it does not form oxides.
2 CO + O2 ==> 2 CO2
x ml x/2 ml
x ml
In the input the amount of O2 present = x/2 ml
Total volume of gas mixture + O2 = 200 + x/2 ml
Total volume of gas mixture + O2 = 200 + x/2 ml
Resulting mixture: total : 200 ml as:
N2: y ml CO2: x + (200 - x - y) = 200 - y ml
Contraction (reduction) in volume of gases is
40 ml = 200 + x/2 - 200 = x/2
x = volume of CO in the mixture = 80 ml
Now the mixture of CO2 + N2 is passed through base KOH. All CO2 reacts and N2
is not reactive with K OH.
2 K OH + CO2 ==> K2 CO3 + H2 O
So volume of gas mixture reduces by 50% of 200 ml ie., 100 ml.
=> 200 - y = 100 ml
=> y = 100 ml
So we had 80 ml of CO, 100 ml of N2, and 20 ml of CO2 in the mixture initially.
Ratio: of volumes of CO2 : CO : N2 = 1 : 4 : 5
Most Upvoted Answer
200 ml of a gaseous mixture containing CO, CO2and N2on complete combus...
Given data:
Volume of gaseous mixture = 200 ml
Contraction observed on complete combustion = 40 ml
Reduction in volume on passing through KOH = 50 %
To find: Volume ratio of VCO2:VCO:VN2 in original mixture
Step 1: Calculation of volume of O2 required for complete combustion
As the given gaseous mixture contains CO, CO2, and N2, we need to calculate the volume of O2 required for complete combustion of each gas separately.
For complete combustion of CO, the balanced chemical equation is:
2CO + O2 → 2CO2
From the equation, 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2
So, 1 mole of CO requires 1/2 mole of O2 for complete combustion
Volume of O2 required for complete combustion of CO = (1/2) × 200 ml = 100 ml
For complete combustion of CO2, the balanced chemical equation is:
CO2 + O2 → 2CO2
From the equation, 1 mole of CO2 reacts with 1 mole of O2 to produce 2 moles of CO2
So, 1 mole of CO2 requires 1 mole of O2 for complete combustion
Volume of O2 required for complete combustion of CO2 = 1 × 200 ml = 200 ml
For complete combustion of N2, the balanced chemical equation is:
N2 + 2O2 → 2NO2
From the equation, 1 mole of N2 reacts with 2 moles of O2 to produce 2 moles of NO2
So, 1 mole of N2 requires 2 moles of O2 for complete combustion
Volume of O2 required for complete combustion of N2 = 2 × 200 ml = 400 ml
Total volume of O2 required for complete combustion of the given gaseous mixture = 100 ml + 200 ml + 400 ml = 700 ml
Step 2: Calculation of volume of CO2 produced on complete combustion
From the balanced chemical equation for complete combustion of CO and CO2, we can see that the volume of CO2 produced will be equal to the volume of CO consumed.
Volume of CO consumed = contraction observed on complete combustion = 40 ml
Volume of CO2 produced on complete combustion = 40 ml
Step 3: Calculation of volume of N2 remaining after complete combustion
The total volume of gases before combustion = 200 ml
Volume of gases consumed in combustion = contraction observed on complete combustion + volume of O2 used for combustion
= 40 ml + 700 ml = 740 ml
Volume of N2 remaining after combustion = Total volume of gases before combustion - Volume of gases consumed in combustion
= 200 ml - 740 ml = -540 ml (negative sign indicates that there is no N2 remaining after combustion)
Step 4: Calculation of volume of gases after reduction by KOH
On passing through KOH solution, the volume of gases is reduced by 50 %
Volume of gases after reduction by KOH = (100 - 50) % of (40 ml + 40 ml + 0 ml)
= (50/100) × 80 ml
= 40 ml
Step 5: Calculation of volume ratio of
Volume of gaseous mixture = 200 ml
Contraction observed on complete combustion = 40 ml
Reduction in volume on passing through KOH = 50 %
To find: Volume ratio of VCO2:VCO:VN2 in original mixture
Step 1: Calculation of volume of O2 required for complete combustion
As the given gaseous mixture contains CO, CO2, and N2, we need to calculate the volume of O2 required for complete combustion of each gas separately.
For complete combustion of CO, the balanced chemical equation is:
2CO + O2 → 2CO2
From the equation, 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2
So, 1 mole of CO requires 1/2 mole of O2 for complete combustion
Volume of O2 required for complete combustion of CO = (1/2) × 200 ml = 100 ml
For complete combustion of CO2, the balanced chemical equation is:
CO2 + O2 → 2CO2
From the equation, 1 mole of CO2 reacts with 1 mole of O2 to produce 2 moles of CO2
So, 1 mole of CO2 requires 1 mole of O2 for complete combustion
Volume of O2 required for complete combustion of CO2 = 1 × 200 ml = 200 ml
For complete combustion of N2, the balanced chemical equation is:
N2 + 2O2 → 2NO2
From the equation, 1 mole of N2 reacts with 2 moles of O2 to produce 2 moles of NO2
So, 1 mole of N2 requires 2 moles of O2 for complete combustion
Volume of O2 required for complete combustion of N2 = 2 × 200 ml = 400 ml
Total volume of O2 required for complete combustion of the given gaseous mixture = 100 ml + 200 ml + 400 ml = 700 ml
Step 2: Calculation of volume of CO2 produced on complete combustion
From the balanced chemical equation for complete combustion of CO and CO2, we can see that the volume of CO2 produced will be equal to the volume of CO consumed.
Volume of CO consumed = contraction observed on complete combustion = 40 ml
Volume of CO2 produced on complete combustion = 40 ml
Step 3: Calculation of volume of N2 remaining after complete combustion
The total volume of gases before combustion = 200 ml
Volume of gases consumed in combustion = contraction observed on complete combustion + volume of O2 used for combustion
= 40 ml + 700 ml = 740 ml
Volume of N2 remaining after combustion = Total volume of gases before combustion - Volume of gases consumed in combustion
= 200 ml - 740 ml = -540 ml (negative sign indicates that there is no N2 remaining after combustion)
Step 4: Calculation of volume of gases after reduction by KOH
On passing through KOH solution, the volume of gases is reduced by 50 %
Volume of gases after reduction by KOH = (100 - 50) % of (40 ml + 40 ml + 0 ml)
= (50/100) × 80 ml
= 40 ml
Step 5: Calculation of volume ratio of
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200 ml of a gaseous mixture containing CO, CO2and N2on complete combus...
Option C is correct...
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200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer?
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200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? 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 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer?.
200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? 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 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer?.
Solutions for 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 200 ml of a gaseous mixture containing CO, CO2and N2on complete combustion in just sufficient amount of O2showed contraction of 40 ml. When the resulting gases were passed through KOH solution it reduces by 50 % then calculate the volume ratio of VCO2:VCO: VN2in original mixture.a)4 : 1 : 5b)2 : 3 : 5c)1 : 4 : 5d)1 : 3 : 5Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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