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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) + H2(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?
Verified Answer
Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of...
C = 84/12 = 7 mole
H2 = 12 g = 6 mole
O2 = 56/22.4 = 5/2 mole
12C + 11H2 + 11/2 O2 → C12H22O11
L.R. = O2
11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)
= 5/11 × 342
= 155.45 g
H2 = 12 g = 6 mole
O2 = 56/22.4 = 5/2 mole
12C + 11H2 + 11/2 O2 → C12H22O11
L.R. = O2
11/2 mole O2 produce 1 mole sucrose 5/2 mole O2 will for 5/11 mole sucrose mass of sucrose = 5/11 × (mol. mass)
= 5/11 × 342
= 155.45 g
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Most Upvoted Answer
Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of...
Given reaction: C(s) + H2(g) + O2(g) → C12H22O11(s)
To find the mass of sucrose produced, we need to first balance the equation:
12C(s) + 11H2(g) + 11O2(g) → C12H22O11(s)
Now, we can use the principle of stoichiometry to find the mass of sucrose produced.
1 mole of sucrose (C12H22O11) = 342 g
From the balanced equation, we can see that 12 moles of carbon (C) react to produce 1 mole of sucrose (C12H22O11). Similarly, 22 moles of hydrogen (H2) and 11 moles of oxygen (O2) are also required to produce 1 mole of sucrose.
So, we need to first find the number of moles of carbon, hydrogen and oxygen present in the given amounts.
Number of moles of carbon = 84 g / 12 g/mol = 7 moles
Number of moles of hydrogen = 12 g / 2 g/mol = 6 moles
Number of moles of oxygen = (1 atm * 56 L) / (0.0821 atm L/mol K * 273 K) = 2 moles
From the balanced equation, we can see that the limiting reactant is oxygen (O2) as we have only 2 moles of it. This means that we can produce only 1/11th of a mole of sucrose.
So, the mass of sucrose produced = (1/11) * 342 g = 31.09 g
Therefore, the correct answer is option B (155.5 g).
To find the mass of sucrose produced, we need to first balance the equation:
12C(s) + 11H2(g) + 11O2(g) → C12H22O11(s)
Now, we can use the principle of stoichiometry to find the mass of sucrose produced.
1 mole of sucrose (C12H22O11) = 342 g
From the balanced equation, we can see that 12 moles of carbon (C) react to produce 1 mole of sucrose (C12H22O11). Similarly, 22 moles of hydrogen (H2) and 11 moles of oxygen (O2) are also required to produce 1 mole of sucrose.
So, we need to first find the number of moles of carbon, hydrogen and oxygen present in the given amounts.
Number of moles of carbon = 84 g / 12 g/mol = 7 moles
Number of moles of hydrogen = 12 g / 2 g/mol = 6 moles
Number of moles of oxygen = (1 atm * 56 L) / (0.0821 atm L/mol K * 273 K) = 2 moles
From the balanced equation, we can see that the limiting reactant is oxygen (O2) as we have only 2 moles of it. This means that we can produce only 1/11th of a mole of sucrose.
So, the mass of sucrose produced = (1/11) * 342 g = 31.09 g
Therefore, the correct answer is option B (155.5 g).
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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) + 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 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) + 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 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) + 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 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) + 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 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) + 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 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) + 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 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) + 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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