How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO4 ?

Question

How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO4 ? (A) 3.54 litre (B) 7.08 litre (C) 1.77 litre (D) None of these

Tardigrade Admin · Accepted Answer

Meq. of Cl2=Meq. of KMnO4 (w/71 / 2)× 1000=(10/31.6)× 1000 So wC l2=11.23 g So VC l2=(22.4 × 11.23/71)=3.54 litre.

Q. How many litres of $C l_{2}$ at STP will be liberated by the oxidation of NaCl with 10 g $K M n O_{4}$ ?

3.54 litre

7.08 litre

1.77 litre

None of these

Meq. of $C l_{2} = M e q .$ of $K M n O_{4}$
$\frac{w}{71/2} \times 1000 = \frac{10}{31.6} \times 1000$
So $w_{C l_{2}} = 11.23$ g
So $V_{C l_{2}} = \frac{22.4 \times 11.23}{71} = 3.54$ litre.

Q. How many litres of Cl2​ at STP will be liberated by the oxidation of NaCl with 10 g KMnO4​ ?