NEET Exam > NEET Questions > 1L of O at STP is made to react with 3L of ca...
Start Learning for Free
1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.?
Most Upvoted Answer
1L of O at STP is made to react with 3L of carbon monoxide at STP . Ca...
Given Data:
- Volume of oxygen (O2) = 1L
- Volume of carbon monoxide (CO) = 3L
- Both gases are at STP
Step 1: Write the balanced chemical equation
2CO + O2 → 2CO2
Step 2: Calculate the moles of each reactant
- Moles of O2 = Volume / 22.4 (at STP) = 1/22.4 = 0.0446 mol
- Moles of CO = Volume / 22.4 = 3/22.4 = 0.134 mol
Step 3: Determine the limiting reagent
- Using the stoichiometry of the balanced equation, we find that 2 moles of CO react with 1 mole of O2.
- Since there is an equal number of moles of CO2 produced for every mole of O2 and CO, the limiting reagent is the reactant that is present in the lowest amount.
- The moles of O2 available are enough to react with 0.0446 x 2 = 0.0892 mol of CO, which is less than the actual moles of CO present.
- Therefore, O2 is the limiting reagent.
Step 4: Calculate the mass of each substance
- Mass of O2 = moles x molar mass = 0.0446 x 32 = 1.43 g
- Mass of CO = moles x molar mass = 0.134 x 28 = 3.75 g
- Mass of CO2 produced = moles x molar mass = 0.0892 x 44 = 3.93 g
Conclusion:
- After the reaction, the mass of O2 remaining will be 0 g, since it is the limiting reagent.
- The mass of CO remaining will be 3.75 - 3.93 = -0.18 g, which indicates that all the CO will be consumed in the reaction.
- The mass of CO2 produced will be 3.93 g.
Therefore, the limiting reagent is O2, and the mass of each substance found after the reaction is 0 g of O2, 0 g of CO, and 3.93 g of CO2.
- Volume of oxygen (O2) = 1L
- Volume of carbon monoxide (CO) = 3L
- Both gases are at STP
Step 1: Write the balanced chemical equation
2CO + O2 → 2CO2
Step 2: Calculate the moles of each reactant
- Moles of O2 = Volume / 22.4 (at STP) = 1/22.4 = 0.0446 mol
- Moles of CO = Volume / 22.4 = 3/22.4 = 0.134 mol
Step 3: Determine the limiting reagent
- Using the stoichiometry of the balanced equation, we find that 2 moles of CO react with 1 mole of O2.
- Since there is an equal number of moles of CO2 produced for every mole of O2 and CO, the limiting reagent is the reactant that is present in the lowest amount.
- The moles of O2 available are enough to react with 0.0446 x 2 = 0.0892 mol of CO, which is less than the actual moles of CO present.
- Therefore, O2 is the limiting reagent.
Step 4: Calculate the mass of each substance
- Mass of O2 = moles x molar mass = 0.0446 x 32 = 1.43 g
- Mass of CO = moles x molar mass = 0.134 x 28 = 3.75 g
- Mass of CO2 produced = moles x molar mass = 0.0892 x 44 = 3.93 g
Conclusion:
- After the reaction, the mass of O2 remaining will be 0 g, since it is the limiting reagent.
- The mass of CO remaining will be 3.75 - 3.93 = -0.18 g, which indicates that all the CO will be consumed in the reaction.
- The mass of CO2 produced will be 3.93 g.
Therefore, the limiting reagent is O2, and the mass of each substance found after the reaction is 0 g of O2, 0 g of CO, and 3.93 g of CO2.
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
|
|
Top Courses for NEETView all
1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.?
Question Description
1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.?.
1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.?.
Solutions for 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? 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 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? defined & explained in the simplest way possible. Besides giving the explanation of
1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.?, a detailed solution for 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? has been provided alongside types of 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? theory, EduRev gives you an
ample number of questions to practice 1L of O at STP is made to react with 3L of carbon monoxide at STP . Calculate the mass of each substance found after the reaction . Which one is the limiting reagent.? 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
|
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