JEE Exam > JEE Questions > At 300 K and 1 atm, 15 mL of a gaseous hydroc...
Start Learning for Free
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]
- a)C4H8
- b)C4H10
- c)C3H6
- d)C3H8
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air...
Volume of O2 used = 
= 75 ml
= 75 ml∴ From the reaction of combustion
1 ml CxHy requires =
So, 4x + y = 20
x = 3
y = 8
x = 3
y = 8
Most Upvoted Answer
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air...
Given data:
- Temperature (T) = 300 K
- Pressure (P) = 1 atm
- Volume of hydrocarbon (V_hydrocarbon) = 15 mL
- Volume of air (V_air) = 375 mL
- Oxygen content in air (O2_air) = 20% by volume
- Final volume of gases after combustion (V_final) = 330 mL
Key equations:
1. Ideal Gas Law: PV = nRT
2. Volume ratio of gases in combustion: V_hydrocarbon : V_oxygen : V_carbon dioxide : V_water = a : b : c : d (where a, b, c, d are coefficients of balanced combustion equation)
Solution:
Step 1: Finding the volume of oxygen (V_oxygen) required for combustion:
Since the oxygen content in air is 20%, the volume of oxygen in the air can be calculated as:
V_oxygen = V_air * O2_air = 375 mL * 0.20 = 75 mL
Step 2: Calculating the volume of carbon dioxide (V_carbon dioxide) produced:
Since the volume of oxygen is half the volume of hydrocarbon, the balanced combustion equation can be written as:
V_hydrocarbon : V_oxygen : V_carbon dioxide : V_water = 1 : 0.5 : c : d
Since the total volume after combustion is 330 mL, we can write:
V_hydrocarbon + V_oxygen + V_carbon dioxide + V_water = 330 mL
Substituting the given values, we get:
15 mL + 75 mL + V_carbon dioxide + V_water = 330 mL
V_carbon dioxide + V_water = 330 mL - 90 mL = 240 mL
Step 3: Calculating the volume of water (V_water) produced:
Since water is in liquid form, its volume doesn't change with temperature and pressure. Therefore, we can write:
V_water = 240 mL
Step 4: Balancing the combustion equation:
Using the volume ratios of the balanced combustion equation, we have:
1 : 0.5 : c : 240 mL
The lowest whole number ratio for c is 480. Therefore, the balanced combustion equation is:
1 : 0.5 : 480 : 240
Step 5: Determining the formula of the hydrocarbon:
The balanced combustion equation is:
C_hydrocarbonH_hydrocarbon + 0.5O2 → 480CO2 + 240H2O
From the equation, we can determine that the hydrocarbon contains 4 carbon atoms (C4H_hydrocarbon). Therefore, the formula of the hydrocarbon is C3H8 (option D).
Conclusion:
The formula of the hydrocarbon is C3H8 (option D).
- Temperature (T) = 300 K
- Pressure (P) = 1 atm
- Volume of hydrocarbon (V_hydrocarbon) = 15 mL
- Volume of air (V_air) = 375 mL
- Oxygen content in air (O2_air) = 20% by volume
- Final volume of gases after combustion (V_final) = 330 mL
Key equations:
1. Ideal Gas Law: PV = nRT
2. Volume ratio of gases in combustion: V_hydrocarbon : V_oxygen : V_carbon dioxide : V_water = a : b : c : d (where a, b, c, d are coefficients of balanced combustion equation)
Solution:
Step 1: Finding the volume of oxygen (V_oxygen) required for combustion:
Since the oxygen content in air is 20%, the volume of oxygen in the air can be calculated as:
V_oxygen = V_air * O2_air = 375 mL * 0.20 = 75 mL
Step 2: Calculating the volume of carbon dioxide (V_carbon dioxide) produced:
Since the volume of oxygen is half the volume of hydrocarbon, the balanced combustion equation can be written as:
V_hydrocarbon : V_oxygen : V_carbon dioxide : V_water = 1 : 0.5 : c : d
Since the total volume after combustion is 330 mL, we can write:
V_hydrocarbon + V_oxygen + V_carbon dioxide + V_water = 330 mL
Substituting the given values, we get:
15 mL + 75 mL + V_carbon dioxide + V_water = 330 mL
V_carbon dioxide + V_water = 330 mL - 90 mL = 240 mL
Step 3: Calculating the volume of water (V_water) produced:
Since water is in liquid form, its volume doesn't change with temperature and pressure. Therefore, we can write:
V_water = 240 mL
Step 4: Balancing the combustion equation:
Using the volume ratios of the balanced combustion equation, we have:
1 : 0.5 : c : 240 mL
The lowest whole number ratio for c is 480. Therefore, the balanced combustion equation is:
1 : 0.5 : 480 : 240
Step 5: Determining the formula of the hydrocarbon:
The balanced combustion equation is:
C_hydrocarbonH_hydrocarbon + 0.5O2 → 480CO2 + 240H2O
From the equation, we can determine that the hydrocarbon contains 4 carbon atoms (C4H_hydrocarbon). Therefore, the formula of the hydrocarbon is C3H8 (option D).
Conclusion:
The formula of the hydrocarbon is C3H8 (option D).
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer?
Question Description
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer?.
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer?.
Solutions for At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer?, a detailed solution for At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL.Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: [JEE M 2016]a)C4H8b)C4H10c)C3H6d)C3H8Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup