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The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:
- a)200 mL of 0.4 N HCl
- b)200 mL of 0.2 N HCl
- c)100 mL of 0.2 N HCl
- d)100 mL of 0.1 N HCl
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH...
1 mol of urea = 2 mol of NH3
60 g of urea = 2 mol of NH3
0.6 g of urea = 2/60 x 0.6mol = 0.02 mol of NH3
mol of NH3 = mol of HCl
∴ mol of HCl = 0.02 mol
⇒ Normality of HC1 = 0.2 N
Volume of HCL = 100 mL
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The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH...
Question Analysis:
The question asks us to determine which solution can neutralize the ammonia (NH3) released from the reaction between urea (NH2CONH2) and sodium hydroxide (NaOH). To answer this question, we need to calculate the amount of ammonia released and compare it to the amount of acid in each solution.
Given Information:
- Mass of urea (NH2CONH2) = 0.6 g
- Reaction: NH2CONH2 + NaOH → NH3 + H2O
- Option a) 200 mL of 0.4 N HCl
- Option b) 200 mL of 0.2 N HCl
- Option c) 100 mL of 0.2 N HCl
- Option d) 100 mL of 0.1 N HCl
Calculating the Amount of Ammonia Released:
1. Calculate the number of moles of urea (NH2CONH2):
- The molar mass of urea (NH2CONH2) = 60 g/mol
- Number of moles of urea = mass / molar mass = 0.6 g / 60 g/mol = 0.01 mol
2. From the balanced equation, we can see that 1 mol of urea produces 1 mol of ammonia (NH3).
Therefore, the number of moles of ammonia released = 0.01 mol
3. Calculate the volume of 0.01 M NH3 released using the Ideal Gas Law:
- The molar volume of an ideal gas at STP = 22.4 L/mol
- Volume of NH3 released = moles × molar volume = 0.01 mol × 22.4 L/mol = 0.224 L = 224 mL
Calculating the Amount of Acid in Each Solution:
1. Option a) 200 mL of 0.4 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.2 L × 0.4 mol/L = 0.08 mol
2. Option b) 200 mL of 0.2 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.2 L × 0.2 mol/L = 0.04 mol
3. Option c) 100 mL of 0.2 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.1 L × 0.2 mol/L = 0.02 mol
4. Option d) 100 mL of 0.1 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.1 L × 0.1 mol/L = 0.01 mol
Comparing the Amount of Acid to the Amount of Ammonia:
- Option a) 0.08 mol of HCl is greater than 0.01 mol of NH3.
- Option b) 0.04 mol of HCl is greater than 0.01 mol of NH3
The question asks us to determine which solution can neutralize the ammonia (NH3) released from the reaction between urea (NH2CONH2) and sodium hydroxide (NaOH). To answer this question, we need to calculate the amount of ammonia released and compare it to the amount of acid in each solution.
Given Information:
- Mass of urea (NH2CONH2) = 0.6 g
- Reaction: NH2CONH2 + NaOH → NH3 + H2O
- Option a) 200 mL of 0.4 N HCl
- Option b) 200 mL of 0.2 N HCl
- Option c) 100 mL of 0.2 N HCl
- Option d) 100 mL of 0.1 N HCl
Calculating the Amount of Ammonia Released:
1. Calculate the number of moles of urea (NH2CONH2):
- The molar mass of urea (NH2CONH2) = 60 g/mol
- Number of moles of urea = mass / molar mass = 0.6 g / 60 g/mol = 0.01 mol
2. From the balanced equation, we can see that 1 mol of urea produces 1 mol of ammonia (NH3).
Therefore, the number of moles of ammonia released = 0.01 mol
3. Calculate the volume of 0.01 M NH3 released using the Ideal Gas Law:
- The molar volume of an ideal gas at STP = 22.4 L/mol
- Volume of NH3 released = moles × molar volume = 0.01 mol × 22.4 L/mol = 0.224 L = 224 mL
Calculating the Amount of Acid in Each Solution:
1. Option a) 200 mL of 0.4 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.2 L × 0.4 mol/L = 0.08 mol
2. Option b) 200 mL of 0.2 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.2 L × 0.2 mol/L = 0.04 mol
3. Option c) 100 mL of 0.2 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.1 L × 0.2 mol/L = 0.02 mol
4. Option d) 100 mL of 0.1 N HCl:
- Number of moles of HCl = volume (L) × concentration (mol/L) = 0.1 L × 0.1 mol/L = 0.01 mol
Comparing the Amount of Acid to the Amount of Ammonia:
- Option a) 0.08 mol of HCl is greater than 0.01 mol of NH3.
- Option b) 0.04 mol of HCl is greater than 0.01 mol of NH3
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The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer?
Question Description
The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer?.
The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by:a)200 mL of 0.4 N HClb)200 mL of 0.2 N HClc)100 mL of 0.2 N HCld)100 mL of 0.1 N HClCorrect answer is option 'C'. Can you explain this answer?.
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