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Calculate the limiting reagent
1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O
2)32g S8+32g 8O2 GIVES 8SO2?
1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O
2)32g S8+32g 8O2 GIVES 8SO2?
Most Upvoted Answer
Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3...
Limiting Reagent Calculation for Reaction 1
To determine the limiting reagent in the reaction:
Reaction:
4 moles Cu + 20 moles HNO3 → 3 moles Cu(NO3)2 + 4 moles H2O
- Molar Ratios:
- Cu : HNO3 = 4 : 20 = 1 : 5
- Available Moles:
- Cu = 4 moles
- HNO3 = 20 moles
- Required Moles of HNO3 for 4 moles Cu:
- Required = 4 moles Cu * (5 moles HNO3 / 1 mole Cu) = 20 moles HNO3
- Conclusion:
- Both reactants are in exact stoichiometric amounts. No limiting reagent here.
Limiting Reagent Calculation for Reaction 2
For the second reaction:
Reaction:
32g S8 + 32g O2 → 8 SO2
- Molar Mass Calculation:
- Molar mass of S8 = 8 * 32 g/mol = 256 g/mol
- Molar mass of O2 = 2 * 16 g/mol = 32 g/mol
- Available Moles:
- Moles of S8 = 32g / 256 g/mol = 0.125 moles
- Moles of O2 = 32g / 32 g/mol = 1 mole
- Molar Ratios:
- S8 : O2 = 1 : 4 (from the balanced equation)
- Required Moles of O2 for 0.125 moles S8:
- Required = 0.125 moles S8 * (4 moles O2 / 1 mole S8) = 0.5 moles O2
- Conclusion:
- Available O2 = 1 mole > Required O2 = 0.5 moles. S8 is the limiting reagent.
In summary, the limiting reagents are:
- Reaction 1: No limiting reagent.
- Reaction 2: S8 is the limiting reagent.
To determine the limiting reagent in the reaction:
Reaction:
4 moles Cu + 20 moles HNO3 → 3 moles Cu(NO3)2 + 4 moles H2O
- Molar Ratios:
- Cu : HNO3 = 4 : 20 = 1 : 5
- Available Moles:
- Cu = 4 moles
- HNO3 = 20 moles
- Required Moles of HNO3 for 4 moles Cu:
- Required = 4 moles Cu * (5 moles HNO3 / 1 mole Cu) = 20 moles HNO3
- Conclusion:
- Both reactants are in exact stoichiometric amounts. No limiting reagent here.
Limiting Reagent Calculation for Reaction 2
For the second reaction:
Reaction:
32g S8 + 32g O2 → 8 SO2
- Molar Mass Calculation:
- Molar mass of S8 = 8 * 32 g/mol = 256 g/mol
- Molar mass of O2 = 2 * 16 g/mol = 32 g/mol
- Available Moles:
- Moles of S8 = 32g / 256 g/mol = 0.125 moles
- Moles of O2 = 32g / 32 g/mol = 1 mole
- Molar Ratios:
- S8 : O2 = 1 : 4 (from the balanced equation)
- Required Moles of O2 for 0.125 moles S8:
- Required = 0.125 moles S8 * (4 moles O2 / 1 mole S8) = 0.5 moles O2
- Conclusion:
- Available O2 = 1 mole > Required O2 = 0.5 moles. S8 is the limiting reagent.
In summary, the limiting reagents are:
- Reaction 1: No limiting reagent.
- Reaction 2: S8 is the limiting reagent.
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Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2?
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Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2? 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 reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2?.
Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2? 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 reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the limiting reagent1)4 mole 3Cu+20 mole 8HNO3 GIVES 3Cu(NO3)2+4H2O2)32g S8+32g 8O2 GIVES 8SO2?.
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