In a water treatment plant, $C l_{2}$ used for the treatment of water is produced from the following reaction:
$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$
If during each feed, $1 L$ $K M n O_{4}$ having $79 %$ (w/v) $K M n O_{4}$ and $9 L H C l$ with $d = 1.825$ g/mL and $10 %$ (w/w) $H C l$ are entered and if that percent yield is $80 %$ , then match the following.

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Solution

Given,
$1 ℓ K M n O_{4}$ having $79$ % $(w / v) \Rightarrow 79 ℊ$ of $K M n O_{4}$ in $100 ㎖$ of solution
$∴ 1 ℓ$ will contain $790 ℊ$ of $K M n O_{4}$ Molecular weight is $158$
$∴ n_{K M n O_{4}} = \frac{790}{158} = 5$ $d_{H C l} = 1.825 ℊ / ㎖ = 1825 ℊ / ℓ$
$∴ 9 ℓ$ of solution $= (9 \times 1825) ℊ = 16425 ℊ$
$10$ % $w / w$ of $H C l \Rightarrow 1642.5 ℊ$ of $H C l$ in $16425 ℊ$ of solution
$∴ n_{H C l} = \frac{1642.5}{36.5} = 45$
$2 K M n O_{4} + 16 H C l ⟶ 2 K C l + 2 M n {C l}_{2} + 8 H_{2} O + 5 {C l}_{2}$
Clearly, $H C l$ is in excess
$n_{{C l}_{2}}$ formed $= 5 \times \frac{5}{2} \times 0.8 = 10 ㏖$
$28.4 ℊ$ of ${C l}_{2}$ heats $1 ℓ$ if water
Total ${C l}_{2}$ formed $= 10 ㏖ = 710 ℊ$
Volume of water that can be treated $= \frac{710}{28.4} = 25 ℓ$
$\frac{V o l . o f w a t e r t r e a t e d}{V o l . o f t o t a l f e e d} = \frac{25}{(1 + 9)} = 2.5$

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