Question

2,Which has greater normality as an oxidizing agent in the presence of acid, a solution containing 0.32g of KMnO4 per 100ml or one containing 0.72g of K2Cr2O7 per 150ml? How many mililiters pf 0.1N stannous salt will 50ml of each of the oxidizing agents react with in the presence of acid to form stannic salt? How many grams of Sn are present in each liter of the stannic salt？

Answer 1

A) Normality:

Normality of KMnO4 in the presence of acid :

In acidic medium, Mn+7 -------> Mn+2

Change in oxidation state = 5

Equivalent weight of KMnO4 = molecular weight/5 = 158/5 = 31.6 Eq

Hence,

Normality of KMnO4 = ( weight/ equivalent weight) x (1/vol in L)

= (0.32 g/ 31.6 Eq) x (1/0.1 L)

= 0.101 N

Normality of K2Cr2O7 in the presence of acid :

In acidic medium, Cr⁺⁶ -------> 2 Cr⁺⁶

Change in oxidation state = 12- 6 = 6

Equivalent weight of K2Cr2O7 = molecular weight/6 = 294/6 = 49 Eq

Hence,

Normality of K2Cr2O7 = ( weight/ equivalent weight) x (1/vol in L)

= (0.72 g/ 49 Eq) x (1/0.15 L)

= 0.098 N

Therefore, KMnO4 has greater normality.

-------------------------------------------------------------------------------

B) How many mililiters pf 0.1N stannous salt will 50ml of each of the oxidizing agents react with in the presence of acid to form stannic salt?

KMnO4:

N1V1 = N2V2

0.101 N x 50 ml = 0.1 N x V2

V2 = 50.5 ml

Therefore, 50.5 ml of 0.1N stannous salt will react with 50ml of KMnO4.

K2Cr2O7:

N1V1 = N2V2

0.098 N x 50 ml = 0.1 N x V2

V2 = 49 ml

Therefore, 49 ml of 0.1 N stannous salt will react with 50ml of K2Cr2O7.