Mole Fraction Question!?
Here's the problem:
From the combustion of C2H6, a mixture of CO2 and H2O with a mass of 5.65 g was produced. The volume of the vessel containing the mixture was 8.45L and the temperature was 150 degrees C.
A)Write balanced equation:
Here I came up with 2C2H6 + 7O2= 4CO2+6H20
B)Determine mole fraction of each:
Answer:
2C2H6 + 7O2 = 4CO2 + 6H20
The equation you gave is balanced correctly.
Now let us calculate the molar balance of the equation
Taken the atomic weight (mass) for each of the molecules in the equation:
2*12*2 +2*6 + 7*16*2 = 4*12 +8*16 + 12 +6*16
48 + 12 + 108 = 48 + 128 + 108
Or 284 gram/mole reacted with 284 gram/mole
This shows that your equation is correctly balanced.
176g/mole of 2C2H6 +108 g/mole of Oxygen reacted with 60g/mole+224g/mole H2O
From the balanced equation we know that:
2 moles of C2H6 + 7 moles of Oxygen reacted to produce 4 moles of Carbon dioxide and 6 moles of water!
The outcome of the combustion produced 5.65 gram of Carbon dioxide and water as specified in your question which show incorrect information!.
The atomic mass of Carbon dioxide is 12 + 32 = 44 gram = one mole
The atomic moss of water is 2 + 16 = 18 gram = one mole
44 gram + 18 gram = 60 grams make one mole of compound mixture of water and of Carbon dioxide
The percentage of Carbon dioxide of the mixture is 44/60 = 0.733
Since we have 5.65 of the mixture produced the reaction, then out of that which was produced we have 0.733 * 5.65 = 4.14 gram of Carbon dioxide and 1.51 of Water
For molar fraction calculation we have: 12+32 =44 grams of Carbon dioxide makes one mole and 18 grams of water makes one mole.
Molar percentage of CO2 would be equivalent to 4.14 /60 = 0.069 moles of Carbon dioxide
1.51 / 224 = 0.00674 moles of water has been produced
The information provided in your question does not seem to be complete
I tried my best to assist. I hope I have shown what you are looking for!
Use Dalton's Law of Partial Pressures.
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From the combustion of C2H6, a mixture of CO2 and H2O with a mass of 5.65 g was produced. The volume of the vessel containing the mixture was 8.45L and the temperature was 150 degrees C.
A)Write balanced equation:
Here I came up with 2C2H6 + 7O2= 4CO2+6H20
B)Determine mole fraction of each:
Answer:
2C2H6 + 7O2 = 4CO2 + 6H20
The equation you gave is balanced correctly.
Now let us calculate the molar balance of the equation
Taken the atomic weight (mass) for each of the molecules in the equation:
2*12*2 +2*6 + 7*16*2 = 4*12 +8*16 + 12 +6*16
48 + 12 + 108 = 48 + 128 + 108
Or 284 gram/mole reacted with 284 gram/mole
This shows that your equation is correctly balanced.
176g/mole of 2C2H6 +108 g/mole of Oxygen reacted with 60g/mole+224g/mole H2O
From the balanced equation we know that:
2 moles of C2H6 + 7 moles of Oxygen reacted to produce 4 moles of Carbon dioxide and 6 moles of water!
The outcome of the combustion produced 5.65 gram of Carbon dioxide and water as specified in your question which show incorrect information!.
The atomic mass of Carbon dioxide is 12 + 32 = 44 gram = one mole
The atomic moss of water is 2 + 16 = 18 gram = one mole
44 gram + 18 gram = 60 grams make one mole of compound mixture of water and of Carbon dioxide
The percentage of Carbon dioxide of the mixture is 44/60 = 0.733
Since we have 5.65 of the mixture produced the reaction, then out of that which was produced we have 0.733 * 5.65 = 4.14 gram of Carbon dioxide and 1.51 of Water
For molar fraction calculation we have: 12+32 =44 grams of Carbon dioxide makes one mole and 18 grams of water makes one mole.
Molar percentage of CO2 would be equivalent to 4.14 /60 = 0.069 moles of Carbon dioxide
1.51 / 224 = 0.00674 moles of water has been produced
The information provided in your question does not seem to be complete
I tried my best to assist. I hope I have shown what you are looking for!
Use Dalton's Law of Partial Pressures.
The answers post by the user, for information only, FunQA.com does not guarantee the right.
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