H2O2 - 2KI 40- yield- I2 - 2KOHH2O2 - 2KMnO4 - 3H2SO450- yield- K2SO4 - 2MnSO4 - 3O2 - 4H2O150 mL of a H2O2 sample was divided into two parts- The first part was treated with KI and the formed KOH required 200mL of M2 H2SO4 for complete neutralisation- The other part was treated with KMnO4 yielding 6-72 L of O2 at 1 atm and 273 K- Using the - yields indicated- find the volume strength of the H2O2 sample used-

Moles of H_2SO_4 = 0.2×12 mol = 0.1 mol Moles of used KOH will be = 0.1× 2 mol =0.2 mol H_2O_2 + 2KI 40% yield⟶ I_2 + 2KOH ∴ Moles of H_2O_2 used in the fir ...

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Byju's Answer

Standard XIII

Chemistry

Introduction to Oxidation and Reduction

H2O2 + 2KI 40...

Question

$H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ I_{2} + 2 K O H H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
$150 mL$ 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 complete neutralisation. 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 $%$ yields indicated, find the volume strength of the $H_{2} O_{2}$ sample used.

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Solution

$Moles of H_{2} S O_{4} = 0.2 \times \frac{1}{2} m o l = 0.1 m o l$
$Moles of used K O H will be = 0.1 \times 2 m o l = 0.2 m o l$

$H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ I_{2} + 2 K O H$
$∴$ Moles of $H_{2} O_{2}$ used in the first reaction
$= \frac{0.2}{2} \times \frac{1}{0.4} = 0.25$
According to the question, the moles of $O_{2}$ produced $= \frac{6.72}{22.4} = 0.3 m o l$

Given,
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
$∴$ Moles of $H_{2} O_{2}$ used in the second reaction
$= \frac{0.3}{3 \times 0.5} = 0.2$
So, the total moles of $H_{2} O_{2}$ consumed $= (0.25 + 0.2) m o l = 0.45 m o l$
Molarity of $H_{2} O_{2} = \frac{0.45}{0.15} = 3 M$
Volume strength $= 11.2 \times 3 = 33.6$

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Similar questions

H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ 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 ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O

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

\frac{M}{2} H_{2} S O_{4}

for complete neutralisation. The other part was treated with

K M n O_{4}

yielding

6.72 L

O_{2}

1 atm

and

273 K

. Using the

%

yields indicated, find the volume strength of the

H_{2} O_{2}

sample used.

H_{2} O_{2} + 2 K I 40 % yield - ---- \to I_{2} + 2 K O H

H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % yield - ---- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O

150 mL

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

M / 2

H_{2} S O_{4}

for neurtralisation. The other part was treated with

K M n O 4

, which yielded 6.74 litre of

O_{2}

at 1 atm and

273 K

. Using the percentage yield indicated, find the volume strength of the

H_{2} O_{2}

sample used.

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

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

$150$ 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.