How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO4 in acidic medium: (Atomic weight: Mn = 55 and K = 39)

Question

How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO4 in acidic medium: (Atomic weight: Mn = 55 and K = 39) (A) 3.54 (B) 7.08 (C) 1.77 (D) None of these

Tardigrade Admin · Accepted Answer

Cl- + MnO4- → Mn+2 + Cl2 Eq of Cl2 = Eq of KMnO4 2[mole of CI2]=5[(10/158)] mole of Cl2 =(50/2×158)=0.15823 mole volume of Cl2 at STP = 0.15823 × 22.4 = 3.54L

Q. How many litres of $C l_{2}$ at $STP$ will be liberated by the oxidation of $N a Cl$ with $10 g K M n O_{4}$ in acidic medium: (Atomic weight $: M n = 55$ and $K = 39$ )

$3.54$

$7.08$

$1.77$

None of these

$C l^{-} + M n O_{4}^{-} \to M n^{+ 2} + C l_{2}$
$Eq o f C l_{2} = Eq o f K M n O_{4}$
$2 [m o l eo f C I_{2}] = 5 [\frac{10}{158}]$
mole of $C l_{2} = \frac{50}{2 \times 158}$ =0.15823 mole
volume of $C l_{2}$ at $STP = 0.15823 \times 22.4 = 3.54 L$

Q. How many litres of Cl2​ at STP will be liberated by the oxidation of NaCl with 10gKMnO4​ in acidic medium: (Atomic weight:Mn=55 and K=39)

3.54

7.08

1.77