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Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2 at 1 atm & 273 K according to given reaction, is C(s) + H 2(g) + O2(g) → C12H22O11(s)
- a)138.5
- b)155.5
- c)172.5
- d)199.5
Correct answer is option 'B'. Can you explain this answer?
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Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of ...
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Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of ...
To find the mass of sucrose produced, we first need to determine the limiting reactant in the reaction between carbon, hydrogen, and oxygen.
1. Calculate the moles of each reactant:
- Moles of carbon: 84 g / 12.01 g/mol = 7.00 mol
- Moles of hydrogen: 12 g / 1.01 g/mol = 11.88 mol
- Moles of oxygen: 56 L x (1 mol/22.4 L) = 2.50 mol
2. Determine the mole ratio of the reactants based on the balanced chemical equation:
C12H22O11 + 12O2 → 12CO2 + 11H2O
The mole ratio of carbon to oxygen is 7.00 mol : 12.50 mol, or approximately 0.56 : 1. The mole ratio of hydrogen to oxygen is 11.88 mol : 12.50 mol, or approximately 0.95 : 1.
3. Identify the limiting reactant:
Since the mole ratio of carbon to oxygen is less than 1, carbon is the limiting reactant. This means that all the carbon will react with the available oxygen, and any excess oxygen will remain unreacted.
4. Calculate the moles of sucrose produced:
The balanced chemical equation shows that 1 mole of sucrose is produced for every 12 moles of oxygen consumed. Since the reaction consumes 12.50 moles of oxygen, it will produce:
1/12 x 12.50 mol = 1.04 mol of sucrose
5. Calculate the mass of sucrose produced:
The molar mass of sucrose C12H22O11 is 342.3 g/mol. Therefore:
1.04 mol x 342.3 g/mol = 356.2 g of sucrose
Therefore, the mass of sucrose produced by mixing 84 g of carbon, 12 g of hydrogen, and 56 L of oxygen at 1 atm is 356.2 g.
1. Calculate the moles of each reactant:
- Moles of carbon: 84 g / 12.01 g/mol = 7.00 mol
- Moles of hydrogen: 12 g / 1.01 g/mol = 11.88 mol
- Moles of oxygen: 56 L x (1 mol/22.4 L) = 2.50 mol
2. Determine the mole ratio of the reactants based on the balanced chemical equation:
C12H22O11 + 12O2 → 12CO2 + 11H2O
The mole ratio of carbon to oxygen is 7.00 mol : 12.50 mol, or approximately 0.56 : 1. The mole ratio of hydrogen to oxygen is 11.88 mol : 12.50 mol, or approximately 0.95 : 1.
3. Identify the limiting reactant:
Since the mole ratio of carbon to oxygen is less than 1, carbon is the limiting reactant. This means that all the carbon will react with the available oxygen, and any excess oxygen will remain unreacted.
4. Calculate the moles of sucrose produced:
The balanced chemical equation shows that 1 mole of sucrose is produced for every 12 moles of oxygen consumed. Since the reaction consumes 12.50 moles of oxygen, it will produce:
1/12 x 12.50 mol = 1.04 mol of sucrose
5. Calculate the mass of sucrose produced:
The molar mass of sucrose C12H22O11 is 342.3 g/mol. Therefore:
1.04 mol x 342.3 g/mol = 356.2 g of sucrose
Therefore, the mass of sucrose produced by mixing 84 g of carbon, 12 g of hydrogen, and 56 L of oxygen at 1 atm is 356.2 g.
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Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. Can you explain this answer?
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Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. 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 Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. 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 Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. Can you explain this answer?.
Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. 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 Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. 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 Mass of sucrose C12H22O11produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2at 1 atm& 273 K according to given reaction, is C(s) + H2(g) + O2(g) →C12H22O11(s)a)138.5b)155.5c)172.5d)199.5Correct answer is option 'B'. Can you explain this answer?.
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