percentage yield

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 )

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%

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!

Really? I got it from this website: http://www.uen.org/utahlink/tours/tourElement.cgi?element_id=42288&tour_id=17891&category_id=33176

Do you know any good websites that have A2 chemistry questions? Just the maths stuff, that's where I'm weak at.