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10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2?
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10 gm of S reacts with excess of O2 to form 15% of so2 find the percen...
Understanding the Reaction
When sulfur (S) reacts with excess oxygen (O2), it forms sulfur dioxide (SO2). The balanced chemical equation for this reaction is:
S + O2 → SO2
Calculating Moles of Sulfur
- Molar mass of sulfur (S) = 32 g/mol
- Moles of sulfur = Mass/Molar mass = 10 g / 32 g/mol = 0.3125 moles
Expected Production of SO2
- From the balanced equation, 1 mole of S produces 1 mole of SO2.
- Therefore, 0.3125 moles of S will produce 0.3125 moles of SO2.
Calculating Actual Yield
- If the reaction produces 15% of SO2, we can calculate the mass of SO2 produced.
- Molar mass of SO2 = 64 g/mol.
- Theoretical yield of SO2 = 0.3125 moles × 64 g/mol = 20 g.
- Actual yield = 15% of 20 g = 3 g.
Calculating Percentage Yield
- Percentage yield = (Actual yield / Theoretical yield) × 100
- Percentage yield = (3 g / 20 g) × 100 = 15%.
Conclusion
The percentage yield of the reaction between sulfur and oxygen forming sulfur dioxide is 15%. This indicates that only a fraction of the expected product was obtained under the given conditions.
When sulfur (S) reacts with excess oxygen (O2), it forms sulfur dioxide (SO2). The balanced chemical equation for this reaction is:
S + O2 → SO2
Calculating Moles of Sulfur
- Molar mass of sulfur (S) = 32 g/mol
- Moles of sulfur = Mass/Molar mass = 10 g / 32 g/mol = 0.3125 moles
Expected Production of SO2
- From the balanced equation, 1 mole of S produces 1 mole of SO2.
- Therefore, 0.3125 moles of S will produce 0.3125 moles of SO2.
Calculating Actual Yield
- If the reaction produces 15% of SO2, we can calculate the mass of SO2 produced.
- Molar mass of SO2 = 64 g/mol.
- Theoretical yield of SO2 = 0.3125 moles × 64 g/mol = 20 g.
- Actual yield = 15% of 20 g = 3 g.
Calculating Percentage Yield
- Percentage yield = (Actual yield / Theoretical yield) × 100
- Percentage yield = (3 g / 20 g) × 100 = 15%.
Conclusion
The percentage yield of the reaction between sulfur and oxygen forming sulfur dioxide is 15%. This indicates that only a fraction of the expected product was obtained under the given conditions.
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10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2?
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10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2?.
10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 gm of S reacts with excess of O2 to form 15% of so2 find the percentage yield of the reaction s+o2gives rise to so2?.
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