NEET Exam > NEET Questions > Calculate the limiting reagent in the given r...
Start Learning for Free
Calculate the limiting reagent in the given reactions
1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O
2) S8+8SO2
32 g 32g?
1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O
2) S8+8SO2
32 g 32g?
Most Upvoted Answer
Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gi...
Limiting Reagent Analysis
To determine the limiting reagent in the given reactions, we need to analyze the stoichiometry of each reaction.
Reaction 1: 3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O
- Molar Mass Calculation:
- Copper (Cu): 63.55 g/mol
- Nitric Acid (HNO3): 63.01 g/mol
- Assumed Amounts:
- Let's say we have 100 g of Cu.
- Moles of Cu = 100 g / 63.55 g/mol ≈ 1.57 moles
- Moles of HNO3 needed = (8/3) * 1.57 ≈ 4.19 moles
- Mass of HNO3 = 4.19 moles * 63.01 g/mol ≈ 264.56 g
- Conclusion for Reaction 1:
- If you have less than 264.56 g of HNO3, then HNO3 is the limiting reagent. If you have more, Cu is limiting.
Reaction 2: S8 + 8 SO2 → 8 S3O2
- Molar Mass Calculation:
- Sulfur (S): 32.07 g/mol
- SO2: 64.07 g/mol
- Assumed Amounts:
- 32 g of S8 = 1 mol (since molar mass of S8 = 256.16 g/mol)
- Moles of SO2 needed = 8 moles (1 mol of S8 requires 8 moles of SO2).
- Conclusion for Reaction 2:
- If you have less than 8 moles of SO2 (which is 8 * 64.07 ≈ 512.56 g), then SO2 is the limiting reagent. If more, S8 is limiting.
Final Remarks:
- Identify the limiting reagent based on available amounts.
- Calculate moles and compare with stoichiometric ratios.
To determine the limiting reagent in the given reactions, we need to analyze the stoichiometry of each reaction.
Reaction 1: 3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O
- Molar Mass Calculation:
- Copper (Cu): 63.55 g/mol
- Nitric Acid (HNO3): 63.01 g/mol
- Assumed Amounts:
- Let's say we have 100 g of Cu.
- Moles of Cu = 100 g / 63.55 g/mol ≈ 1.57 moles
- Moles of HNO3 needed = (8/3) * 1.57 ≈ 4.19 moles
- Mass of HNO3 = 4.19 moles * 63.01 g/mol ≈ 264.56 g
- Conclusion for Reaction 1:
- If you have less than 264.56 g of HNO3, then HNO3 is the limiting reagent. If you have more, Cu is limiting.
Reaction 2: S8 + 8 SO2 → 8 S3O2
- Molar Mass Calculation:
- Sulfur (S): 32.07 g/mol
- SO2: 64.07 g/mol
- Assumed Amounts:
- 32 g of S8 = 1 mol (since molar mass of S8 = 256.16 g/mol)
- Moles of SO2 needed = 8 moles (1 mol of S8 requires 8 moles of SO2).
- Conclusion for Reaction 2:
- If you have less than 8 moles of SO2 (which is 8 * 64.07 ≈ 512.56 g), then SO2 is the limiting reagent. If more, S8 is limiting.
Final Remarks:
- Identify the limiting reagent based on available amounts.
- Calculate moles and compare with stoichiometric ratios.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g?
Question Description
Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g?.
Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g?.
Solutions for Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? defined & explained in the simplest way possible. Besides giving the explanation of
Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g?, a detailed solution for Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? has been provided alongside types of Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? theory, EduRev gives you an
ample number of questions to practice Calculate the limiting reagent in the given reactions1) 3 cu +8HNO3 gives 3Cu (NO3)2+2NO+4H2O2) S8+8SO2 32 g 32g? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup