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2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33?
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2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is m...
Given:
- 2 moles of N2
- 1 mole of H2
- Reaction: N2(g) + 3H2(g) → 2NH3(g)
- NH3 is neutralized with 0.1 M H2SO4
To find:
The volume of H2SO4 required to neutralize NH3.
Solution:
Step 1: Calculate the moles of NH3 produced.
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Since we have 2 moles of N2, the number of moles of NH3 produced will be:
2 moles N2 × (2 moles NH3 / 1 mole N2) = 4 moles NH3
Step 2: Calculate the volume of 0.1 M H2SO4 required to neutralize 4 moles of NH3.
The balanced equation between NH3 and H2SO4 is:
2NH3 + H2SO4 → (NH4)2SO4
From the equation, it can be seen that 2 moles of NH3 react with 1 mole of H2SO4.
Therefore, 4 moles of NH3 will react with (4/2) = 2 moles of H2SO4.
Now, we can calculate the volume of 0.1 M H2SO4 required using the formula:
Moles of solute = Molarity × Volume of solution in liters
2 moles H2SO4 = (0.1 M) × Volume of H2SO4 in liters
Volume of H2SO4 in liters = 2 moles H2SO4 / (0.1 M) = 20 L
Step 3: Convert the volume of H2SO4 from liters to milliliters.
1 L = 1000 mL
Therefore, the volume of H2SO4 required in milliliters = 20 L × 1000 mL/L = 20000 mL
Step 4: Convert the volume of H2SO4 from milliliters to liters.
1 mL = 0.001 L
Therefore, the volume of H2SO4 required in liters = 20000 mL × 0.001 L/mL = 20 L
Answer:
The volume of H2SO4 required to neutralize NH3 is 20 liters.
Option (4) 3.33 L is incorrect.
- 2 moles of N2
- 1 mole of H2
- Reaction: N2(g) + 3H2(g) → 2NH3(g)
- NH3 is neutralized with 0.1 M H2SO4
To find:
The volume of H2SO4 required to neutralize NH3.
Solution:
Step 1: Calculate the moles of NH3 produced.
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Since we have 2 moles of N2, the number of moles of NH3 produced will be:
2 moles N2 × (2 moles NH3 / 1 mole N2) = 4 moles NH3
Step 2: Calculate the volume of 0.1 M H2SO4 required to neutralize 4 moles of NH3.
The balanced equation between NH3 and H2SO4 is:
2NH3 + H2SO4 → (NH4)2SO4
From the equation, it can be seen that 2 moles of NH3 react with 1 mole of H2SO4.
Therefore, 4 moles of NH3 will react with (4/2) = 2 moles of H2SO4.
Now, we can calculate the volume of 0.1 M H2SO4 required using the formula:
Moles of solute = Molarity × Volume of solution in liters
2 moles H2SO4 = (0.1 M) × Volume of H2SO4 in liters
Volume of H2SO4 in liters = 2 moles H2SO4 / (0.1 M) = 20 L
Step 3: Convert the volume of H2SO4 from liters to milliliters.
1 L = 1000 mL
Therefore, the volume of H2SO4 required in milliliters = 20 L × 1000 mL/L = 20000 mL
Step 4: Convert the volume of H2SO4 from milliliters to liters.
1 mL = 0.001 L
Therefore, the volume of H2SO4 required in liters = 20000 mL × 0.001 L/mL = 20 L
Answer:
The volume of H2SO4 required to neutralize NH3 is 20 liters.
Option (4) 3.33 L is incorrect.
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2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33?
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2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33? 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 2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33?.
2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33? 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 2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 moles of N2 is mixed with 1 mole of H2 and the reaction mixture is made to undergo NH3 formation according to given reaction: N2(g) 3H2(g) : 2NH3(g) NH3 produced is neutralized with 0.1 M H2SO4 . Calculate the volume(in l) of H2SO4 required. (1)6.67 (2)0.03 (3)0.06 (4)3.33?.
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