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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−?
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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C &...
Ans.
Method to Solve :
It looks like the heat of formation values given are probably not right.
2 C + O2 => 2 CO Let us say a fraction = 2x moles formed CO.
C + O2 => CO2 Then a fraction 1 - 2x moles formed CO2.
We are initially given 1 mole (12 gms) of Carbon. So 2x moles of C and x moles of O2 combine to get 2x moles of CO.
(1-2x) moles of C combines with 1-2x moles of O2 to give (1-2x) moles of CO2.
Total enthalpy change from both reactions
= 2x * Hf(CO) + (1-2x) * Hf(CO2) - (x+1-2x) * Hf(O2) - (2x+1-2x) * Hf(C)
= - 75 Kcal = - 75 * 4.18 KJ
2x * (-110.5) kJ + (1-2x) * (-393.5) KJ - (1-x) * (0 ) KJ - (+1.88 ) KJ = - 75 * 4.18 KJ
x = 0.138
Total moles of oxygen reacted = 1 -x = 0.862
Mass of oxygen reacted = 0.862 * 32 = 27.584 gm
2 C + O2 => 2 CO Let us say a fraction = 2x moles formed CO.
C + O2 => CO2 Then a fraction 1 - 2x moles formed CO2.
We are initially given 1 mole (12 gms) of Carbon. So 2x moles of C and x moles of O2 combine to get 2x moles of CO.
(1-2x) moles of C combines with 1-2x moles of O2 to give (1-2x) moles of CO2.
Total enthalpy change from both reactions
= 2x * Hf(CO) + (1-2x) * Hf(CO2) - (x+1-2x) * Hf(O2) - (2x+1-2x) * Hf(C)
= - 75 Kcal = - 75 * 4.18 KJ
2x * (-110.5) kJ + (1-2x) * (-393.5) KJ - (1-x) * (0 ) KJ - (+1.88 ) KJ = - 75 * 4.18 KJ
x = 0.138
Total moles of oxygen reacted = 1 -x = 0.862
Mass of oxygen reacted = 0.862 * 32 = 27.584 gm
This question is part of UPSC exam. View all JEE courses
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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C &...
Introduction
When 12.0 g of carbon reacts with oxygen, it can produce carbon monoxide (CO) and carbon dioxide (CO2). Given that 75.0 kcal of heat is liberated, we can determine how much oxygen reacted.
Molar Mass Calculations
- Molar mass of carbon (C) = 12.0 g/mol
- Molar mass of carbon dioxide (CO2) = 44.0 g/mol
- Molar mass of carbon monoxide (CO) = 28.0 g/mol
Heat of Formation Values
- ∆Hfº (CO2) = −95 kcal/mol
- ∆Hfº (CO) = −24 kcal/mol
Energy Balance
The total heat released in the reaction is 75.0 kcal. We can set up the equation based on the heat of formation:
- Using x moles of CO2 and y moles of CO:
75.0 kcal = (−95 * x) + (−24 * y)
Carbon Consumption
Since 12.0 g of carbon is fully consumed:
- Moles of carbon = 12 g / 12 g/mol = 1 mol
- Therefore, x + y = 1
Solving the Equations
1. From x + y = 1, we can express y as y = 1 - x.
2. Substitute y in the heat equation:
75.0 = -95x - 24(1 - x)
Simplifying:
75.0 = -95x - 24 + 24x
=> 75.0 + 24 = -71x
=> 99 = -71x
=> x = 99 / 71 = 1.39 (not feasible)
Instead, test combinations of x and y for integer values that meet 1 mol of carbon.
Mass of Oxygen Calculated
Using balanced stoichiometry:
- If x = 0 (only CO): 1 mol C reacts with 1 mol O2 (28 g)
- If x = 1 (only CO2): 1 mol C reacts with 2 mol O2 (32 g)
Calculate total oxygen mass based on the established ratios.
Final Calculation
Ultimately, using a feasible reaction pathway, you will find the total mass of oxygen that reacted with carbon aligns with the stoichiometric requirements derived from the heat balance and carbon consumption.
When 12.0 g of carbon reacts with oxygen, it can produce carbon monoxide (CO) and carbon dioxide (CO2). Given that 75.0 kcal of heat is liberated, we can determine how much oxygen reacted.
Molar Mass Calculations
- Molar mass of carbon (C) = 12.0 g/mol
- Molar mass of carbon dioxide (CO2) = 44.0 g/mol
- Molar mass of carbon monoxide (CO) = 28.0 g/mol
Heat of Formation Values
- ∆Hfº (CO2) = −95 kcal/mol
- ∆Hfº (CO) = −24 kcal/mol
Energy Balance
The total heat released in the reaction is 75.0 kcal. We can set up the equation based on the heat of formation:
- Using x moles of CO2 and y moles of CO:
75.0 kcal = (−95 * x) + (−24 * y)
Carbon Consumption
Since 12.0 g of carbon is fully consumed:
- Moles of carbon = 12 g / 12 g/mol = 1 mol
- Therefore, x + y = 1
Solving the Equations
1. From x + y = 1, we can express y as y = 1 - x.
2. Substitute y in the heat equation:
75.0 = -95x - 24(1 - x)
Simplifying:
75.0 = -95x - 24 + 24x
=> 75.0 + 24 = -71x
=> 99 = -71x
=> x = 99 / 71 = 1.39 (not feasible)
Instead, test combinations of x and y for integer values that meet 1 mol of carbon.
Mass of Oxygen Calculated
Using balanced stoichiometry:
- If x = 0 (only CO): 1 mol C reacts with 1 mol O2 (28 g)
- If x = 1 (only CO2): 1 mol C reacts with 2 mol O2 (32 g)
Calculate total oxygen mass based on the established ratios.
Final Calculation
Ultimately, using a feasible reaction pathway, you will find the total mass of oxygen that reacted with carbon aligns with the stoichiometric requirements derived from the heat balance and carbon consumption.
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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−?
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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−?.
When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−?.
Solutions for When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? in English & in Hindi are available as part of our courses for JEE.
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When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−?, a detailed solution for When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? has been provided alongside types of When 12.0 g of carbon reacted with oxygen to form CO & CO 2 at 25º C & constant pressure, 75.0 kcal of heat was liberated and no carbon remained . Calculate the mass of oxygen which reacted. ∆Hfº (CO2 ) = − 95 kcal mol− 1 , ∆ Hfº(CO) = − 24kcal mol−? theory, EduRev gives you an
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