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3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)
- a)2 Moles of C2I-16 left for combustion
- b)Volume of C02 at S.T.P. produced after combustion 44.8 litre.
- c)Volume of water produced is 54 ml
- d)None
Correct answer is option 'B,C'. Can you explain this answer?
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Verified Answer
3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 ...
Moles of C2H6 = 3
moles of C2H6 mixed = 60/30 = 2
total mole of C2H6 = 5
moles removed = 
moles of C2 H6 left =1
Now, C2 H6 + 7/2 O2 → 2CO2 + 3H2O
Clearly 3 moles of H2O or 54 gm H2O will be formed
volume of H2O = 54 ml
Most Upvoted Answer
3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 ...
To solve this problem, we need to follow a step-by-step approach. Let's break it down into smaller parts.
1. Calculate the number of moles of C2H6 initially present:
Given that 3 moles of C2H6 are mixed with 60 g of the gas, we can use the formula:
Number of moles = mass / molar mass
Molar mass of C2H6 = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol
Number of moles of C2H6 = 60 g / 30.07 g/mol = 1.995 moles (approx. 2 moles)
2. Calculate the number of molecules of C2H6 initially present:
Given that 2.4 x 10^24 molecules of the gas are removed, we can use Avogadro's number to calculate the initial number of molecules:
Number of molecules = number of moles x Avogadro's number
Number of molecules = 2 moles x 6 x 10^23 molecules/mole = 1.2 x 10^24 molecules
3. Calculate the moles of C2H6 left after removing 2.4 x 10^24 molecules:
Given that 2.4 x 10^24 molecules of the gas are removed, we can calculate the remaining moles using the formula:
Number of moles = number of molecules / Avogadro's number
Number of moles = (1.2 x 10^24 molecules) / (6 x 10^23 molecules/mole) = 2 moles
4. Calculate the moles of C2H6 that undergo combustion:
Since 2 moles of C2H6 are initially present and 2 moles remain after removing the molecules, we can conclude that all 2 moles undergo combustion.
5. Calculate the volume of CO2 produced after combustion at STP:
1 mole of C2H6 produces 2 moles of CO2 during combustion. Therefore, 2 moles of C2H6 will produce 4 moles of CO2.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters. So, 4 moles of CO2 will occupy a volume of:
Volume of CO2 = 4 moles x 22.4 liters/mole = 89.6 liters
6. Calculate the volume of water produced:
During the combustion of C2H6, water is produced as a byproduct. The balanced equation for the combustion of C2H6 is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
From the equation, we can see that for every 2 moles of C2H6 combusted, 3 moles of water are produced.
Since all 2 moles of C2H6 undergo combustion, the volume of water produced can be calculated as:
Volume of water = 3 moles x 22.4 liters/mole = 67.2 liters
So, the correct answers are:
b) The volume of CO2 produced after combustion at STP is 44.8 liters.
c) The volume of water produced is 54 ml.
1. Calculate the number of moles of C2H6 initially present:
Given that 3 moles of C2H6 are mixed with 60 g of the gas, we can use the formula:
Number of moles = mass / molar mass
Molar mass of C2H6 = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol
Number of moles of C2H6 = 60 g / 30.07 g/mol = 1.995 moles (approx. 2 moles)
2. Calculate the number of molecules of C2H6 initially present:
Given that 2.4 x 10^24 molecules of the gas are removed, we can use Avogadro's number to calculate the initial number of molecules:
Number of molecules = number of moles x Avogadro's number
Number of molecules = 2 moles x 6 x 10^23 molecules/mole = 1.2 x 10^24 molecules
3. Calculate the moles of C2H6 left after removing 2.4 x 10^24 molecules:
Given that 2.4 x 10^24 molecules of the gas are removed, we can calculate the remaining moles using the formula:
Number of moles = number of molecules / Avogadro's number
Number of moles = (1.2 x 10^24 molecules) / (6 x 10^23 molecules/mole) = 2 moles
4. Calculate the moles of C2H6 that undergo combustion:
Since 2 moles of C2H6 are initially present and 2 moles remain after removing the molecules, we can conclude that all 2 moles undergo combustion.
5. Calculate the volume of CO2 produced after combustion at STP:
1 mole of C2H6 produces 2 moles of CO2 during combustion. Therefore, 2 moles of C2H6 will produce 4 moles of CO2.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters. So, 4 moles of CO2 will occupy a volume of:
Volume of CO2 = 4 moles x 22.4 liters/mole = 89.6 liters
6. Calculate the volume of water produced:
During the combustion of C2H6, water is produced as a byproduct. The balanced equation for the combustion of C2H6 is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
From the equation, we can see that for every 2 moles of C2H6 combusted, 3 moles of water are produced.
Since all 2 moles of C2H6 undergo combustion, the volume of water produced can be calculated as:
Volume of water = 3 moles x 22.4 liters/mole = 67.2 liters
So, the correct answers are:
b) The volume of CO2 produced after combustion at STP is 44.8 liters.
c) The volume of water produced is 54 ml.
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3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer?
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3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer?.
3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer?.
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3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer?, a detailed solution for 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? has been provided alongside types of 3 moles of the gas C2H6is mixed with 60 gm of this gas and 2.4 x 1024 molecules of the gas is removed. The left over gas is combusted in the presence of excess oxygen then : (NA = 6 x 1023) (Density of water = 1 gm/ml)a)2 Moles of C2I-16 left for combustionb)Volume of C02 at S.T.P. produced after combustion 44.8 litre.c)Volume of water produced is 54 mld)NoneCorrect answer is option 'B,C'. Can you explain this answer? theory, EduRev gives you an
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