H2O2-2KII2-2KOH H2O2-2KMnO4-3H2SO4 K2SO4-2MnSO4-3O2-4H2O 150 mL of H2O2 sample was divided into two parts- First part was treated with KI and formed KOH required 200 mL of M2 H2SO4 for neutralization- Other part was treated with KMnO4 for neutralization yielding 6-74 L of O2 at 1 atm and 273 K- Using - yield indicated- find volume strength of H2O2 sample used-

The correct option is A 33.6Mole of H_2SO_4=0.1 Mole of KOH=0.2 Mole of H_2O_2 used in first reaction =0.22×10.4=0.25 Mole of O_2 produced =6.7422.4=0.3 ∴ Mole ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

H2O2+2KII2+2K...

Question

$H_{2} O_{2} + 2 K I 40 % y i e l d - ---- \to I_{2} + 2 K O H$
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d - ---- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
$150 m L$ of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with $K I$ and formed $K O H$ required $200 m L$ of $\frac{M}{2} H_{2} S O_{4}$ for neutralization. Other part was treated with $K M n O_{4}$ for neutralization yielding $6.74 L$ of $O_{2}$ at $1 a t m$ and $273 K$ . Using $%$ yield indicated, find volume strength of $H_{2} O_{2}$ sample used:

5.04

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10.08

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3.36

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33.6

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Solution

The correct option is D $33.6$
Mole of $H_{2} S O_{4} = 0.1$
Mole of $K O H = 0.2$
Mole of $H_{2} O_{2}$ used in first reaction $= \frac{0.2}{2} \times \frac{1}{0.4} = 0.25$
Mole of $O_{2}$ produced $= \frac{6.74}{22.4} = 0.3$
$∴$ Mole of $H_{2} O_{2}$ used in second reaction $= \frac{0.3}{3 \times 0.5} = 0.2$
Total mole of consumed $H_{2} O_{2} = 0.45$
Molarity of $H_{2} O_{2} = \frac{0.45}{0.15} = 3 M$
Volume strength = $11.2 \times 3 = 33.6$ Volumes

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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.