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Balancing Redox Equations

H2O2 +2KI I2+...

Question

$H_{2} O_{2} + 2 K I 40 % yield - ---- \to I_{2} + 2 K O H$
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % yield - ---- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
$150 mL$ of $H_{2} O_{2}$ sample was divided into two parts. The first part was treated with $K I$ and the formed $K O H$ required $200 mL$ of $M / 2$ $H_{2} S O_{4}$ 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.

A

5.04

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B

10.08

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C

3.36

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D

33.6

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Solution

The correct option is D 33.6
$H_{2} O_{2} + 2 K I 40 % - - \to I_{2} + K O H$
$(n \times n_{f})_{K O H} = (n \times n_{f})_{H_{2} S O_{4}}$
$(n \times n_{f})_{K O H} = (M o l a r i t y \times v o l^{m} \times n_{f})_{H_{2} S O_{4}}$
Actual number of mole of $K O H$
$n_{K O H} = 0.2$ actual
$% y i e l d = \frac{a c t u a l}{t h e o r e t i c a l} \times 100$
If mole of x mole of $H_{2} O_{2}$ react it produce 2x mole of $K O H$ ,on putting % yiel formula we get
$x = 0.25$
For
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % yield - ---- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
number of mole of $O_{2}$ produced
$= \frac{6.74}{22.4} = 0.3$
$% y i e l d = \frac{a c t u a l}{t h e o r e t i c a l} \times 100$
If y mole of $H_{2} O_{2}$ produced 3y mole of $K O H$
y=0.2
total number of $H_{2} O_{2} = 0.45$
$M o l a r i t y = \frac{0.45}{0.15} = 3$
volume strength $= 3 \times 11.2 = 33.6$

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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 m L

H_{2} O_{2}

sample was divided into two parts. First part was treated with

K I

K O H

200 m L

\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

O_{2}

1 a t m

273 K

%

yield indicated, find volume strength of

H_{2} O_{2}

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

K O H

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}

273 K

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

H_{2} O_{2}

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

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

$150$ 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.

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

H_{2} O_{2}

sample was divided into two parts. The first part was treated with

K I

K O H

200 mL

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

for complete neutralisation. The other part was treated with

K M n O_{4}

6.72 L

O_{2}

1 atm

273 K

%

yields indicated, find the volume strength of the

H_{2} O_{2}

H_{2} O_{2}

sample was divided into two parts. The first part was treated with

K I

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 %)

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