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)