percentage yield- I'm so confused

A mixture of 480 g of iodine and 600 cm3 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 separated out. The sulphur was filtered off and the solution was purified by fractional distillation. A fraction of HI(aq) was collected containing 440 g of HI in a total volume of 750 cm3. Determine the percentage yield of hydroiodic acid.

My working out:
(mol of I2)
n= m/M = 480/126.9 x 2 = 1.89

(mol of HI)
n = m/M = 440/127.9 = 3.44

% yield = 3.44 x 2/1.89 x 100 = 364.02%


The working out in the mark scheme:
amount I2 reacted = 1.89 mol / HI formed = 3.44 mol (1)
theoretical amount HI produced = 3.78 mol/484 g (1) (<-- why did they multiply the mol of I2 by 2, shouldn't it be the mol of HI they multiply by 2?)
% yield = = 91.0 % (1)

The mark scheme is correct, the maximum yield is twice the moles of I₂ as the amount of iodine is the limiting factor.
Also a 364.02% yield is impossible

A mixture of 480 g of iodine and 600 cm3 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 separated out. The sulphur was filtered off and the solution was purified by fractional distillation. A fraction of HI(aq) was collected containing 440 g of HI in a total volume of 750 cm3. Determine the percentage yield of hydroiodic acid.

My working out:
(mol of I2)
n= m/M = 480/126.9 x 2 = 1.89

(mol of HI)
n = m/M = 440/127.9 = 3.44

% yield = 3.44 x 2/1.89 x 100 = 364.02%


The working out in the mark scheme:
amount I2 reacted = 1.89 mol / HI formed = 3.44 mol (1)
theoretical amount HI produced = 3.78 mol/484 g (1) (<-- why did they multiply the mol of I2 by 2, shouldn't it be the mol of HI they multiply by 2?)
% yield = = 91.0 % (1)

"why did they multiply the mol of I2 by 2, shouldn't it be the mol of HI they multiply by 2"

because percentage yield = mass of useful product / mass of reactants x100
you know that you have 1.89mol of iodine... which REACTS at a 1:1 ratio with H2S...

*see equation*I2 + H2S => 2HI + S

soooo the mass of your reactants = 1.89 + 1.89 = 3.78

so the calculation you need to do is...
(3.44/3.78) x100 = 91.0 percent

So basically you added the moles of BOTH reactants together? Aren't we supposed to find use only one of the reactants (the one that's limiting the reaction)?

So basically you added the moles of BOTH reactants together? Aren't we supposed to find use only one of the reactants (the one that's limiting the reaction)?

1 mole of I₂ goes to 2 moles of HI, so moles of HI = 2 x moles of I₂ . . .

1 mol of I2 produces 2 moles of HI, so to find the theroretical yield of HI you have to multiply by 2? Is that it?

Never mind