Percentage yield help

Hey, its been a while since I have done percentage yield and I got stuck on this one.

7. Hydroiodic acid, HI(aq), is a strong acid that is an aqueous solution of hydrogen iodide. In the laboratory, hydroiodic acid can be prepared by the method below.

A mixture of 480 g of iodine and 600cm^3 of water was put into a flask. The mixture was stirred and hydrogen sulphide gas, H2S(g) was bubbled through for several hours.
The mixture became yellow as sulphur seperated out. The sulphur was filtered off and the solution was purified by fractional distillation. A fraction of HI(aq) was collected containing 440g of HI in a total volume of 750cm^3


(ii)Determine the percentage yield of hydroiodic acid.


I worked out the amount of moles of the HI and I2. I got the moles to be 1.89 for I2 and HI to be 3.44. I then used the ratio to work out the theoretical amount of moles for HI, ratio 1:2. 1.89 multiplied by 2 = 3.78. Then 3.44/3.78 times 100 = 91.0%

Is this right, I'm not that confident on percentage yield.

Set out the equation:

I₂ + H₂S --> 2HI + S

From the description the hydrogen sulphide is added in excess and the iodine is then the limiting reagent:

moles of iodine = 480/254 = 1.89 mol

From the equation this makes 2 x 1.89 mol HI = 3.78 mol = 3.78 x 128 = 483.8g

This represent the theoretical maximum yield (100%)

Percentage yield = 100 x actual yield/theoretical = 90.9%