Question

10cc H2O2 solution when reacted with KI (solution) produced 0.5g iodine. What is percentage purity of H2O2 ?

(1) 66.9%

(2) 0.699%

(3) 100%

(4) Can't be predicted.

{THE CORRECT ANSWER GIVEN IN THE BOOK IS (2) 0.699%}

Answer 1

Chemical Reaction

H2O2 + 2 I- ---> I2 + 2OH-


In 0.5 g of I2 produced, the moles of I2 will be = 0.5127×2=0.00197 mol

Since 1 mol of I2 is formed by 1 mol of H2O2.
0.00197 mol of ​I2 is formed by 0.00197 mol of H2O2, and the mass of H2O2 reacted = 0.00197 x 34 = 0.067 g

Now, to calculate the percentage purity we must know how much grams of H2O2 is present in the solution, which is missing in the question.


percentage purity = Actual amount of H2O2 reacted (g)/Amount of H2O2 used (g)×100 %

Answer 2

Answer:

0.699%

Explanation:

H2O2 + 2 I- ---> I2 + 2OH-

In 0.5 g of I2 produced, the moles of I2 will be = 0.5127×2=0.00197 mol

Since 1 mol of I2 is formed by 1 mol of H2O2.

0.00197 mol of ​I2 is formed by 0.00197 mol of H2O2, and the mass of H2O2 reacted = 0.00197 x 34 = 0.067 g

density of h2o2=1.45=m/v

m=14.5

no of moles=14.5/34

no of moles reacting=0.00197

purity %=