how many littres of cl2 ar stp will be liberated by the oxidation of NACl with 10kg KMnO4
Answers
Answered by
19
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
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
Answered by
2
Answer:3.54 litres.
Explanation:
10×5/158=0.316
Vol of cl2 is 0.316x11.2
=3.54L
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