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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.?
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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.
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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.?
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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.?.
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