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Limiting Reagent or Reactant

Bottle A cont...

Question

Bottle (A) contains $320$ mL of $H_{2} O_{2}$ solution and is labeled with $10 V H_{2} O_{2}$ and bottle (B) contains $80$ mL of $H_{2} O_{2}$ having normality $5 N$ . If bottle (A) and (B) are mixed and solution filled in bottle (C), select the correct label for bottle (C) in terms of volume strength and in terms of g/L.

13.6 V

and

41.285

g/L

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11.2 V

and

0.68

g/L

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5.6 V

and

0.68

g/L

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5.6 V

and

41.285

g/L

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Solution

The correct option is A $13.6 V$ and $41.285$ g/L
$2$ moles of $H_{2} O_{2}$ produce $1$ mole of $O_{2}$ .

$1$ mole of $O_{2}$ occupies $22.4$ litres at STP which is formed from $2 \times 34 = 68$ g/mol (molecular weight of $H_{2} O_{2}$ )

So, if a solution produces $10 V$ of $O_{2}$ , then moles of $H_{2} O_{2}$ is $\frac{10 \times 2}{22.4}$ per litre.

Moles of $H_{2} O_{2}$ in solution B is $\frac{5}{2} = 2.5$

If $2$ solutions of same compound are mixed, then $M_{1} V_{1} + M_{2} V_{2} = M_{3} (V_{1} + V_{2})$

$M_{3} = \frac{M_{1} V_{1} + M_{2} V_{2}}{(V_{1} + V_{2})}$

$M_{3} = \frac{\frac{10 \times 2}{22.4} \times 0.32 + 2.5 \times 0.08}{0.08 + 0.32} = 1.21$

$M_{3} = 1.21$

Amount of $H_{2} O_{2}$ in the new solution is $1.21 \times 34$ which is approximately $= 41.285$ g

$M_{3}$ is $1.21$ and so, volume strength of new solution is $\frac{1.21}{2} \times 22.4$ which is $13.6$ liters.

Hence, the correct option is $A$

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Bottle (A) contains 320 mL of H2O2 solution and is labeled with 10VH2O2 and bottle (B) contains 80 mL of H2O2 having normality 5 N. If bottle (A) and (B) are mixed and solution filled in bottle (C), select the correct label for bottle (C) in terms of volume strength and in terms of g/L.