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Some Basic Concepts of Chemistry

Stoichiometry and Stoichiometric calculations

Hydrogen sulphide reacts with chloric acid in an aqueous solut

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Medium

Solving time: 3 mins

Hydrogen sulphide reacts with chloric acid in an aqueous solution liberating chlorine by the following reaction: $H_{2} S + HClO_{3} \to Cl_{2} + H_{2} SO_{4} + H_{2} O$ In an experiment, $40 mL$ of solution of chloric acid was allowed to react with sufficient amount of $H_{2} S$ . $Cl_{2}$ formed were reduced by $20 mL H_{2} O_{2}$ by the following reaction: $H_{2} O_{2} + Cl_{2} \to HCl + O_{2} ↑$ In the above reaction, $896 mL_{2}$ of $O_{2}$ was evolved at 1 atm and $273 K$ . Which of the following option is incorrect for aforementioned experiment?

Normality of

H_{2} O_{2}

solution used is

4 N

Molarity of original

HClO_{3}

solution used is

2 M

Equivalents of

H_{2} S

reacted with

HClO_{3}

0.05

Moles of

Cl_{2}

produced is

0.04

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Book:

Physical Chemistry I for IIT JEE Main and Advanced (McGraw Hill)

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Harper Discussed

Hydrogen sulphide reacts with chloric acid in an aqueous solution liberating chlorine by the following reaction:

H_{2} S + HClO_{3} \to Cl_{2} + H_{2} SO_{4} + H_{2} O

In an experiment,

40 mL

of solution of chloric acid was allowed to react with sufficient amount of

H_{2} S

Cl_{2}

formed were reduced by

20 mL H_{2} O_{2}

by the following reaction:

H_{2} O_{2} + Cl_{2} \to HCl + O_{2} ↑

In the above reaction,

896 mL_{2}

O_{2}

was evolved at 1 atm and

273 K

. Which of the following option is incorrect for aforementioned experiment?

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

H_{2} S

reacted with

HClO_{3}

0.05

.
(c)

N_{H_{2} O_{2}} = \frac{V}{5.6}

V =

volume of

O_{2}

evolved at STP from

1 L H_{2} O_{2}

solution Moles of

Cl_{2} =

moles of

O_{2} = \frac{896}{22400} = 0.04

equivalent of H_{2} S = equivalent of HClO_{3} = equivalent of Cl_{2} = n_{f} \cdot N = 10 \times 0.04 = 0.4 H_{2} S + 8 HClO_{3} ⟶ 4 Cl_{2} + H_{2} SO_{4} + H_{2} O

No of moles of

HClO_{3} = \frac{1}{2}

moles of

Cl_{2}

= \frac{1}{2} \times 0.04 = 0.02 moles M = \frac{0.02}{40 \times 1 0 ^{- 3}} = \frac{20}{40} = \frac{1}{2} M = 0.5 M

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Question Text	Hydrogen sulphide reacts with chloric acid in an aqueous solution liberating chlorine by the following reaction: $H_{2} S + HClO_{3} \to Cl_{2} + H_{2} SO_{4} + H_{2} O$ In an experiment, $40 mL$ of solution of chloric acid was allowed to react with sufficient amount of $H_{2} S$ . $Cl_{2}$ formed were reduced by $20 mL H_{2} O_{2}$ by the following reaction: $H_{2} O_{2} + Cl_{2} \to HCl + O_{2} ↑$ In the above reaction, $896 mL_{2}$ of $O_{2}$ was evolved at 1 atm and $273 K$ . Which of the following option is incorrect for aforementioned experiment?
Topic	Some Basic Concepts of Chemistry
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	83

Text solutionVerified

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