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A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4?
Most Upvoted Answer
A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H...
C is correct answer.
C+O2-->CO2
0.33/44 mole
1mole of C with 1 mole of O2 gives 1 mole of CO2
to produce 0.33/44=0.03/4 mole,same mole of C is required.
2 H2 + O2-->2 H2O
2× 0.18/18 mole=0.02mole
similarly,0.02 mole of H2 is required
according to law of reciprocal proportion,
C/H ratio =(0.03/4)/0.02=3/8
empirical formula will be C3H8
C+O2-->CO2
0.33/44 mole
1mole of C with 1 mole of O2 gives 1 mole of CO2
to produce 0.33/44=0.03/4 mole,same mole of C is required.
2 H2 + O2-->2 H2O
2× 0.18/18 mole=0.02mole
similarly,0.02 mole of H2 is required
according to law of reciprocal proportion,
C/H ratio =(0.03/4)/0.02=3/8
empirical formula will be C3H8
Community Answer
A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H...
Given Data:
- CO2 produced: 0.33g
- H2O produced: 0.18g
Solution:
Step 1: Calculate moles of CO2 and H2O
- Moles of CO2 = 0.33g / 44g/mol = 0.0075 mol
- Moles of H2O = 0.18g / 18g/mol = 0.01 mol
Step 2: Determine moles of C and H in hydrocarbon
- Moles of C = Moles of CO2 = 0.0075 mol
- Moles of H = 2 * Moles of H2O = 2 * 0.01 mol = 0.02 mol
Step 3: Calculate the ratio of C and H
- Ratio of C to H = 0.0075 mol / 0.02 mol = 0.375
Step 4: Find the empirical formula
- The closest whole number ratio for C:H is 1:3
- Therefore, the empirical formula is C3H8
Conclusion:
- The empirical formula of the hydrocarbon is C3H8.
- CO2 produced: 0.33g
- H2O produced: 0.18g
Solution:
Step 1: Calculate moles of CO2 and H2O
- Moles of CO2 = 0.33g / 44g/mol = 0.0075 mol
- Moles of H2O = 0.18g / 18g/mol = 0.01 mol
Step 2: Determine moles of C and H in hydrocarbon
- Moles of C = Moles of CO2 = 0.0075 mol
- Moles of H = 2 * Moles of H2O = 2 * 0.01 mol = 0.02 mol
Step 3: Calculate the ratio of C and H
- Ratio of C to H = 0.0075 mol / 0.02 mol = 0.375
Step 4: Find the empirical formula
- The closest whole number ratio for C:H is 1:3
- Therefore, the empirical formula is C3H8
Conclusion:
- The empirical formula of the hydrocarbon is C3H8.
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A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4?
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A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4?.
A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrocarbon on complete combustion give 0.33g of CO2 and 0.18 g of H2O. The empirical formula is A) C3H6 B) C2H6 C) C3H8 D) CH4?.
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