Question

how many littres of cl2 ar stp will be liberated by the oxidation of NACl with 10kg KMnO4

Answer 1

Dear frnd... !!!

H+

10 NaCl + 2 KMnO4 ----------> 5 Cl2 + side products

2X 158=316 g 5 x 22.4 at S.T.P.

316 g KMnO4 gives Cl2= 5 x 22.4

10 g KMnO4 gives Cl2= 5 x 22.4 x 10/316

= 3.5 litre

Answer 2

Answer:3.54 litres.

Explanation:

10×5/158=0.316

Vol of cl2 is 0.316x11.2

=3.54L