Answer to Question #17090 in Chemistry for laika

Question

a chemist mixed 100 ml of 10%(w/v) hydroiodic acid, 100 ml of 2 molar sulfuric acid and 40 grams of potassium permanganate. how much molecular iodine will be produced?

Answer 1

So we have 100 * 0.1 =10 g of HI it is 10/128=0.078125 mol ,we also have 100*2/1000= 0.2 mol of H2SO4 and 40/158=0.25316 mole ,as can we see H2SO4 and KMnO4 is in excess.

& 0.078125 m
& 10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
10 5*127

& m=4.96 g