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Stoichiometric Calculations

H2O2+2KI I2+...

Question

$H_{2} O_{2} + 2 K I 40 % y i e l d - ----- \to I_{2} + 2 K O H$
$H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} 50 % y i e l d - ----- \to K_{2} S O_{4} + 2 M n S O_{4} + 3 O_{2} + 4 H_{2} O$
$150$ 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.

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 B $33.6$
Moles of $H_{2} S O_{4} = 0.1$
Moles of $K O H = 0.2$
Moles of $H_{2} O_{2}$ used in first reaction $= \frac{0.2}{2} \times \frac{1}{0.4} = 0.25$
Moles of produced $O_{2} = \frac{6.74}{22.4} = 0.3$
$∴$ Moles of $H_{2} O_{2}$ used in the second reaction $= \frac{0.3}{3 \times 0.5} = 0.2$
Total moles of consumed $H_{2} O_{2} = 0.45$
Molarity of $H_{2} O_{2} = \frac{0.45}{0.15} = 3 M$
Volume strength $= 11.2 \times 3 = 33.6$ V

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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}

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 %)

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} + 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 % 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}

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