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88.33gm of $117.6 %$ oleum is added to 1 litre of $3 {MK}_{4} [Fe (CN)_{6}]$ and further 2 litre of 33.6 volume strength (at $1 atm, 273 K) H_{2} O_{2}$ is added to it resulting in the formation of
$K_{3} [Fe (CN)_{8}]$ as per reaction :
$\begin{aligned} K_{4} [Fe (CN)_{6}] + H_{2} {SO}_{4} + H_{2} O_{2} \\ ⟶ K_{3} [Fe (CN)_{6}] + K_{2} {SO}_{4} + H_{2} O \end{aligned}$
If final volume of the solution is 3 litre, then select the correct statement(s) :

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$H_{2} {SO}_{4}$ is limiting reagent

2 moles of $K_{3} Fe (CN)_{6}$ is formed

If 2ml of resulting solution is heated then $37.3 ml$ of $O_{2}$ at $1 atm, 273 K$ will be evolved from $H_{2} O_{2}$

If 3ml of resulting solution is heated then $37.3 ml$ of $O_{2}$ at $1 atm, 273 K$ will be evolved from $H_{2} O_{2}$

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} n_{H_{2} {SO}_{4}} \\ n_{K_{4} Pe ({CN}_{6}} = 1 \times 3; \\ n_{H_{2} O_{2}} = 2 \times \frac{33.6}{11.2} = 6 \end{array}$

$\begin{array}{l} 2 K_{4} [Fe (CN)_{6}] + H_{2} {SO}_{4} + H_{2} O_{2} \\ (LR) \\ \to 2 K_{3} [Fe (CN)_{6}] + K_{2} {SO}_{4} + 2 H_{2} O \\ 1 \times 2 \end{array}$

$H_{2} O_{2}$ left = 5 moles; $H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$

(c) $\begin{array}{l} \Rightarrow \frac{5}{3} \times 2 \frac{5}{3} \times 2 \times \frac{1}{2} \times 22.4 ml \\ = 37.3 ml \end{array}$

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