150 mL of a $H_{2} O_{2}$ sample was divided into two parts. The first part was treated with $K I$ and the formed $K O H$ required 200 mL of $\frac{M}{2}$ $H_{2} S O_{4}$ for neutralization. The other part was treated with $K M n O_{4}$ , yielding 6.72 L of $O_{2}$ at 1 atm and 273 K. Using the given % yields find the molarity of $H_{2} O_{2}$ .
$H_{2} O_{2} + 2 K I ⟶ I_{2} + 2 K O H (Y i e l d = 40 %)$
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O (Y i e l d = 50 %)$

1.4 M

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3.0 M

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1.5 M

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2.4 M

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Solution

The correct option is B 3.0 M
$M o l e s o f H_{2} S O_{4} = m o l a r i t y \times v o l u m e o f s o l u t i o n (i n l i t r e s) = 0.1$
$M o l e s o f K O H = 2 \times M o l e s o f H_{2} S O_{4} = 2 \times 0.1 = 0.2$
Moles of $H_{2} O_{2}$ used in the first reaction, $= \frac{M o l e s o f K O H}{(2 \times y i e l d o f f i r s t r e a c t i o n)} = \frac{0.2}{2 \times 0.4} = 0.25$

Mole of $O_{2}$ produced $= \frac{v o l u m e o f O_{2} a t S T P}{22.4} = \frac{6.72}{22.4} = 0.3$

Moles of $H_{2} O_{2}$ used in the second reaction,
$= \frac{M o l e s o f O_{2}}{3 \times y i e l d o f s e c o n d r e a c t i o n} = \frac{0.3}{3 \times 0.5} = 0.2$

Total moles of $H_{2} O_{2} = 0.25 + 0.2 = 0.45$

Molarity of $H_{2} O_{2} = \frac{N o . o f m o l e s w a t e r}{V o l u m e o f s o l u t i o n i n l i t r e s} = \frac{0.45}{0.15} = 3 M$

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Similar questions

$H_{2} O_{2} + 2 K I 40 % y i e l d - ----- \to I_{2} + 2 K O H$

$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d - ----- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$

$150$ mL of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with $K I$ and so formed $K O H$ required $200$ mL of $\frac{M}{2} H_{2} S O_{4}$ for neutralisation. Other part was treated with $K M n O_{4}$ yielding $6.74$ litres of $O_{2}$ at $1$ atm and $273$ K. Using $%$ yield indicated, find the volume strength of $H_{2} O_{2}$ sample used.