Class 12 Exam > Class 12 Questions > 40% of a mixture of 0.2 mole of N2 and 0.6 mo...
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40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g) ⇌2NH3 (g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:
- a)4 : 5
- b)5 : 4
- c)7 : 10
- d)8 : 5
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH...
Total mixture is 0.8;40% of it reacts, i.e,
reacts to give 
mole of NH3or 2a=0.16
∴ a=0.08
∴ Initial mole=0.8
Final mole = (0.2-0.08)+(0.6-0.24)+0.16
= 0.12+0.36+0.16=0.64
Ratio of final to initial mole = 0.64/0.8 = 0.8 = 4/5
∴ Initial mole=0.8
Final mole = (0.2-0.08)+(0.6-0.24)+0.16
= 0.12+0.36+0.16=0.64
Ratio of final to initial mole = 0.64/0.8 = 0.8 = 4/5
Most Upvoted Answer
40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH...
Given:
- Mixture contains 0.2 mole of N2 and 0.6 mole of H2
- 40% of the mixture reacts to form NH3
To find:
- The ratio of the final volume to the initial volume of gases
First, let's calculate the moles of N2 and H2 in the mixture that will react.
- 40% of the mixture reacts, so we have:
- Moles of N2 reacting = 0.4 * 0.2 mole = 0.08 mole
- Moles of H2 reacting = 0.4 * 0.6 mole = 0.24 mole
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
So, the moles of NH3 formed will be:
- Moles of NH3 = (0.08 mole of N2) * (2 moles of NH3 / 1 mole of N2) = 0.16 mole
Now, let's calculate the moles of N2 and H2 remaining after the reaction.
- Moles of N2 remaining = 0.2 mole - 0.08 mole = 0.12 mole
- Moles of H2 remaining = 0.6 mole - 0.24 mole = 0.36 mole
We know that 1 mole of any gas occupies the same volume at constant temperature and pressure. Therefore, the ratio of the final volume to the initial volume of gases can be determined by comparing the moles of gases.
The initial moles of gases were:
- Initial moles of N2 = 0.2 mole
- Initial moles of H2 = 0.6 mole
The final moles of gases are:
- Final moles of N2 = 0.12 mole
- Final moles of H2 = 0.36 mole
- Final moles of NH3 = 0.16 mole
Now, let's calculate the ratio of the final volume to the initial volume of gases:
- Final volume = Final moles of N2 + Final moles of H2 + Final moles of NH3
- Initial volume = Initial moles of N2 + Initial moles of H2
Plugging in the values, we get:
- Final volume = 0.12 mole + 0.36 mole + 0.16 mole = 0.64 mole
- Initial volume = 0.2 mole + 0.6 mole = 0.8 mole
Therefore, the ratio of the final volume to the initial volume of gases is:
- Final volume : Initial volume = 0.64 mole : 0.8 mole = 4 : 5
Hence, the correct answer is option 'A' (4 : 5).
- Mixture contains 0.2 mole of N2 and 0.6 mole of H2
- 40% of the mixture reacts to form NH3
To find:
- The ratio of the final volume to the initial volume of gases
First, let's calculate the moles of N2 and H2 in the mixture that will react.
- 40% of the mixture reacts, so we have:
- Moles of N2 reacting = 0.4 * 0.2 mole = 0.08 mole
- Moles of H2 reacting = 0.4 * 0.6 mole = 0.24 mole
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
So, the moles of NH3 formed will be:
- Moles of NH3 = (0.08 mole of N2) * (2 moles of NH3 / 1 mole of N2) = 0.16 mole
Now, let's calculate the moles of N2 and H2 remaining after the reaction.
- Moles of N2 remaining = 0.2 mole - 0.08 mole = 0.12 mole
- Moles of H2 remaining = 0.6 mole - 0.24 mole = 0.36 mole
We know that 1 mole of any gas occupies the same volume at constant temperature and pressure. Therefore, the ratio of the final volume to the initial volume of gases can be determined by comparing the moles of gases.
The initial moles of gases were:
- Initial moles of N2 = 0.2 mole
- Initial moles of H2 = 0.6 mole
The final moles of gases are:
- Final moles of N2 = 0.12 mole
- Final moles of H2 = 0.36 mole
- Final moles of NH3 = 0.16 mole
Now, let's calculate the ratio of the final volume to the initial volume of gases:
- Final volume = Final moles of N2 + Final moles of H2 + Final moles of NH3
- Initial volume = Initial moles of N2 + Initial moles of H2
Plugging in the values, we get:
- Final volume = 0.12 mole + 0.36 mole + 0.16 mole = 0.64 mole
- Initial volume = 0.2 mole + 0.6 mole = 0.8 mole
Therefore, the ratio of the final volume to the initial volume of gases is:
- Final volume : Initial volume = 0.64 mole : 0.8 mole = 4 : 5
Hence, the correct answer is option 'A' (4 : 5).
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40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer?
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40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer?.
40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer?.
Solutions for 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 40% of a mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation, N2(g) + 3h2(g)⇌2NH3(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is:a)4 : 5b)5 : 4c)7 : 10d)8 : 5Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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