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Haber Bosch Process

H 2 O 2 +2KI ...

Question

$H_{2} O_{2} + 2 K I 40 % y i e l d ⟶ I_{2} + 2 K O H$
$5 H_{2} O_{2} + 2 K M n O_{4} + 3 H_{2} {S O}_{4} 50 % y i e l d ⟶$
$K_{2} {S O}_{4} + {2 M n S O}_{4} + 50_{2} + {8 H}_{2} O$
850 mL of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with Kl and formed KOH, required 200 mL of M/2 $H_{2} {S O}_{4}$ for neutralization. Other part was treated with $K M n O_{4}$ yielding 6.74 litre of $O_{2}$ at NTP. Using % yield indicated find volume strength of $H_{2} O_{2}$ sample used:

A

5.94

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B

33.6

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C

3.36

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D

11.2

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Solution

The correct option is A 5.94
$\begin{matrix} Moles of H_{2} {S O}_{4} = 0.1 Moles of K O H = 0.2 Moles of H_{2} O_{2} used in first reaction = \frac{0.2}{2} \times \frac{1}{0.4} = 0.25 \end{matrix}$
$Moles of O_{2} produced = \frac{6 \cdot 74}{22 \cdot 4} = 0.3$
$\begin{matrix} - Moles of H_{2} O_{2} used in the second reaction = \frac{0.3}{3 \times 0.5} = 0.2 \end{matrix}$
$\begin{matrix} Total moles of H_{2} O_{2} consumed \begin{matrix} = 0.45 Molarity of H_{2} O_{2} = \frac{0.45}{0.85} = 0.53 M \end{matrix} \end{matrix}$
$\begin{matrix} Volume strength & = 11.2 \times 0.53 = 5.93 \end{matrix}$

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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}

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

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

$150$ 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 % 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}

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 % 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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