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2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?
- a)60
- b)30
- c)70
- d)35
Correct answer is option 'B'. Can you explain this answer?
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2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields...
Given:
Mass of CO + CO2 = 2 gm
Mass of I2 = 2.54 gm
To find: Mass percentage of CO2 in the original mixture.
Let's assume the mass of CO in the mixture is x gm, then the mass of CO2 will be (2 - x) gm.
The balanced chemical equation for the reaction between CO and I2O5 is:
CO + I2O5 → CO2 + I2
From the equation, we can see that 1 mole of CO reacts with 1 mole of I2O5 to produce 1 mole of CO2 and 1 mole of I2.
1 mole of CO is equal to its molar mass (28 g) and 1 mole of CO2 is equal to its molar mass (44 g). Therefore, the mole ratio of CO to CO2 is 28:44, which simplifies to 7:11.
Now, let's calculate the number of moles of CO and CO2 in the mixture:
Number of moles of CO = Mass of CO / Molar mass of CO
Number of moles of CO = x / 28
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2
Number of moles of CO2 = (2 - x) / 44
According to the given information, the reaction produces 2.54 gm of I2. The molar mass of I2 is 253.8 g/mol. Therefore, the number of moles of I2 can be calculated as:
Number of moles of I2 = Mass of I2 / Molar mass of I2
Number of moles of I2 = 2.54 / 253.8
From the balanced chemical equation, we know that 1 mole of CO reacts with 1 mole of I2 to produce 1 mole of I2. Therefore, the number of moles of CO in the reaction is equal to the number of moles of I2.
Setting up the mole ratio equation:
Number of moles of CO / Number of moles of CO2 = 7 / 11
(x / 28) / ((2 - x) / 44) = 7 / 11
Simplifying the above equation, we get:
11x = 56 - 7x
18x = 56
x = 56 / 18
x = 3.11 gm
Therefore, the mass of CO2 in the original mixture is (2 - x) = (2 - 3.11) = -1.11 gm.
Since the mass of CO2 cannot be negative, it means the assumption made earlier is incorrect. The mass of CO in the mixture cannot be more than 2 gm.
Hence, the correct assumption is that the mass of CO in the mixture is 2 gm, and the mass of CO2 is 0 gm.
Therefore, the mass percentage of CO2 in the original mixture is (0 / 2) * 100 = 0%.
Therefore, the correct answer is option B) 0.
Mass of CO + CO2 = 2 gm
Mass of I2 = 2.54 gm
To find: Mass percentage of CO2 in the original mixture.
Let's assume the mass of CO in the mixture is x gm, then the mass of CO2 will be (2 - x) gm.
The balanced chemical equation for the reaction between CO and I2O5 is:
CO + I2O5 → CO2 + I2
From the equation, we can see that 1 mole of CO reacts with 1 mole of I2O5 to produce 1 mole of CO2 and 1 mole of I2.
1 mole of CO is equal to its molar mass (28 g) and 1 mole of CO2 is equal to its molar mass (44 g). Therefore, the mole ratio of CO to CO2 is 28:44, which simplifies to 7:11.
Now, let's calculate the number of moles of CO and CO2 in the mixture:
Number of moles of CO = Mass of CO / Molar mass of CO
Number of moles of CO = x / 28
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2
Number of moles of CO2 = (2 - x) / 44
According to the given information, the reaction produces 2.54 gm of I2. The molar mass of I2 is 253.8 g/mol. Therefore, the number of moles of I2 can be calculated as:
Number of moles of I2 = Mass of I2 / Molar mass of I2
Number of moles of I2 = 2.54 / 253.8
From the balanced chemical equation, we know that 1 mole of CO reacts with 1 mole of I2 to produce 1 mole of I2. Therefore, the number of moles of CO in the reaction is equal to the number of moles of I2.
Setting up the mole ratio equation:
Number of moles of CO / Number of moles of CO2 = 7 / 11
(x / 28) / ((2 - x) / 44) = 7 / 11
Simplifying the above equation, we get:
11x = 56 - 7x
18x = 56
x = 56 / 18
x = 3.11 gm
Therefore, the mass of CO2 in the original mixture is (2 - x) = (2 - 3.11) = -1.11 gm.
Since the mass of CO2 cannot be negative, it means the assumption made earlier is incorrect. The mass of CO in the mixture cannot be more than 2 gm.
Hence, the correct assumption is that the mass of CO in the mixture is 2 gm, and the mass of CO2 is 0 gm.
Therefore, the mass percentage of CO2 in the original mixture is (0 / 2) * 100 = 0%.
Therefore, the correct answer is option B) 0.
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Community Answer
2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields...
Let in the given mix. Mass of CO = x gm
I2O5+5CO ⟶ 5CO2+I2
1 mole I2 is produced from 5 moles of CO
=0.01 mole I2 is produced from 0.05 moles of CO.
I2O5+5CO ⟶ 5CO2+I2
1 mole I2 is produced from 5 moles of CO
=0.01 mole I2 is produced from 0.05 moles of CO.
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2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. Can you explain this answer?
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2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. 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 2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. Can you explain this answer?.
2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. 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 2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?a)60b)30c)70d)35Correct answer is option 'B'. Can you explain this answer?.
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