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percentage yield watch

1. Hydriodic acid, HI(aq), is a strong acid that is an aqeous solution of hydrogen iodide. In the laboratory, hydriodic acid can be prepared by the method below. A mixture of 480g of iodine and 600 cm3 of water was put into a flask. the mixture was stirred and hydrogen sulfide gas, H2S(g), was bubbled through for several hours. The mixture become yellow as sulfur separated out. The sulfur was filtered off and the solution was purified by fractional distillation. A fraction was collected containing 440g of HI in a total volume of 750 cm3. Determine the percentage yield of hydriodic acid.

This is what the mark scheme says, which I can not understand at all:
amount of I2 reacted = 1.89 mol
Therefore amount of HI formed = 3.44
Theoretical amount of HI produced = 3.78 mol
% yield =3.44 × 100/3.78= 91.0%

This is what I did:
mol of I2: n= m/M = 480/253.8 = 1.89
mol of HI: n/ m/M = 440/127.9 = 3.44
1.89/3.44 x 100 = 55%

I think I understand where I went wrong, this is what I should've done (can someone please check?):
mol of I2: n= m/M = 480/253.8 = 1.89
mol of HI: n= m/M = 440/(2 x 127.9) = 1.72
1.72/1.89 x 100 = 91%
2. (Original post by celina10)
Hydriodic acid, HI(aq), is a strong acid that is an aqeous solution of hydrogen iodide. In the laboratory, hydriodic acid can be prepared by the method below. A mixture of 480g of iodine and 600 cm3 of water was put into a flask. the mixture was stirred and hydrogen sulfide gas, H2S(g), was bubbled through for several hours. The mixture become yellow as sulfur separated out. The sulfur was filtered off and the solution was purified by fractional distillation. A fraction was collected containing 440g of HI in a total volume of 750 cm3. Determine the percentage yield of hydriodic acid.

This is what the mark scheme says, which I can not understand at all:
amount of I2 reacted = 1.89 mol
Therefore amount of HI formed = 3.44
Theoretical amount of HI produced = 3.78 mol
% yield =3.44 × 100/3.78= 91.0%

This is what I did:
mol of I2: n= m/M = 480/106 = 4.53
mol of HI: n/ m/M = 440/54 = 8.148
8.148/4.53 x 100 = 179.87 %
You are dividing the mass of iodine by the incorrect value (which should be relative mass = 254 )
3. (Original post by charco)
You are dividing the mass of iodine by the incorrect value (which should be relative mass = 254 )
whoops, this is what happens when you attempt chemistry at 12 in the night. Ok I changed it now, but I still don't get 91%
4. (Original post by celina10)
Hydriodic acid, HI(aq), is a strong acid that is an aqeous solution of hydrogen iodide. In the laboratory, hydriodic acid can be prepared by the method below. A mixture of 480g of iodine and 600 cm3 of water was put into a flask. the mixture was stirred and hydrogen sulfide gas, H2S(g), was bubbled through for several hours. The mixture become yellow as sulfur separated out. The sulfur was filtered off and the solution was purified by fractional distillation. A fraction was collected containing 440g of HI in a total volume of 750 cm3. Determine the percentage yield of hydriodic acid.

This is what the mark scheme says, which I can not understand at all:
amount of I2 reacted = 1.89 mol
Therefore amount of HI formed = 3.44
Theoretical amount of HI produced = 3.78 mol
% yield =3.44 × 100/3.78= 91.0%

This is what I did:
mol of I2: n= m/M = 480/253.8 = 1.89
mol of HI: n/ m/M = 440/127.9 = 3.44
1.89/3.44 x 100 = 55%

I think I understand where I went wrong, this is what I should've done (can someone please check?):
mol of I2: n= m/M = 480/253.8 = 1.89
mol of HI: n= m/M = 440/(2 x 127.9) = 1.72
1.72/1.89 x 100 = 91%
480g of iodine = 480/254 mol = 1.89 mol

This would make twice as many moles of HI = 3.78 mol

Mr of HI = 128, therefore mass = 483.8g

This represents the 100%

Expt. mass = 440g

Percentage yield = 100 * 440/483.8 = 91%
5. (Original post by charco)
480g of iodine = 480/254 mol = 1.89 mol

This would make twice as many moles of HI = 3.78 mol

Mr of HI = 128, therefore mass = 483.8g

This represents the 100%

Expt. mass = 440g

Percentage yield = 100 * 440/483.8 = 91%
Thank you Do you mind helping me with one more question?
Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the chemical equation below. When 2.00 x 103 g of CaCO3 are heated, the actual yield of CaO is 1.50 x 103 g. What is the percentage yield?

CaCO3 => CaO + CO2

This is what I did:
mol of CaCO3: n= 2 x 103/100.1 = 19.98
mol of CaO: n= 1.5 x 103/56.1 = 26.74
26.74/19.98 x 100 = 133.83%

But the answer is actually 93.8%
6. (Original post by celina10)
Thank you Do you mind helping me with one more question?
Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the chemical equation below. When 2.00 x 103 g of CaCO3 are heated, the actual yield of CaO is 1.50 x 103 g. What is the percentage yield?

CaCO3 => CaO + CO2

This is what I did:
mol of CaCO3: n= 2 x 103/100.1 = 19.98
mol of CaO: n= 1.5 x 103/56.1 = 26.74
26.74/19.98 x 100 = 133.83%

But the answer is actually 93.8%
Moles of calcium carbonate = 2.00 x 103/100 = 20 mol

So theoretical 100% = 20 mol also

BUT, moles of CaO = 1.50 x 103/56 = 26.27 mol .... which is clearly impossible.

There is something wrong with the question as written!
7. (Original post by charco)
Moles of calcium carbonate = 2.00 x 103/100 = 20 mol

So theoretical 100% = 20 mol also

BUT, moles of CaO = 1.50 x 103/56 = 26.27 mol .... which is clearly impossible.

There is something wrong with the question as written!
Really? I got it from this website: http://www.uen.org/utahlink/tours/to...egory_id=33176

Do you know any good websites that have A2 chemistry questions? Just the maths stuff, that's where I'm weak at.
8. (Original post by celina10)
Really? I got it from this website: http://www.uen.org/utahlink/tours/to...egory_id=33176

Do you know any good websites that have A2 chemistry questions? Just the maths stuff, that's where I'm weak at.
Websites are written by humans ... they have errors.

There are past paper exams on www.alevelchem.com

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