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Stoichiometry

The solution ...

Question

The solution which liberates sulphur dioxide gas, from sodium sulphite :

Sodium hydroxide solution

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A weak acid

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Dilute sulphuric acid

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hydrochloric acid

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Solution

The correct option is C Dilute sulphuric acid
Sodium sulphite reacts with dilute sulphuric acid in the manner shown below:
$N a_{2} S O_{3} (s) + H_{2} S O_{4} (l) \to N a_{2} S O_{4} + S O_{2} (g) + H_{2} O$
So, $S O_{2}$ is liberated as a product in the above reaction.

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Similar questions

Q. Solution A is a sodium hydroxide solution. Solution B is a weak acid. C is dilute sulphuric acid. Which solution will
(i) liberate sulphur dioxide from sodium sulphite,
(ii) give a white precipitate with zinc sulphate solution,
(iii) contain solute molecules and ions?

Q. Choose the correct answer from the options given below:
(i) Hydroxide of this metal is soluble in sodium hydroxide solution.
(a) Magnesium
(b) Lead
(c) Silver
(d) Copper

(ii) When dilute sulphuric acid reacts with iron sulphide, the gas evolved is ________.
(a) Hydrogen sulphide
(b) Sulphur dioxide
(c) Sulphur trioxide
(d) Vapour of sulphuric acid.

Q. Given below are word equations. Convert them into chemical equations using symbols and valencies. Balance the equations. Use appropriate signs for gases and precipitates in the reaction.
i) Zinc Sulphide

+

Oxygen gas

\to

Zinc oxide solid

+

Sulphur dioxide gas
ii) Silver nitrate solution

+

Sodium chloride solution

\to

Silver chloride precipitate

+

Sodium nitrate solution
iii) Sulphur solid

+

Concentrated nitric acid

\to

Sulphuric acid

+

Nitrogen dioxide gas

+

Water
iv) Barium chloride Solution

+

Potassium sulphate Solution

\to

Barium sulphate Precipitate

+

Potassium chloride solution
v) Silver nitrate on heating

\to

silver metal

+

nitrogen dioxide gas

+

Oxygen gas.
vi) Aluminium hydroxide

+

Nitric acid

\to

Aluminium nitrate

+

Water.

Write the balanced chemical equations of the following reactions.

(a) Sodium hydroxide + sulphuric acid $\to$ sodium sulphate + water

(b) Potassium bicarbonate + sulphuric acid $\to$ potassium sulphate + carbon dioxide + water

(d) Chlorine + sulphur dioxide + water $\to$ sulphuric acid + hydrogen chloride

(e) Silver nitrate $\to$ silver + nitrogen dioxide + oxygen

(f) Copper + nitric acid $\to$ copper nitrate + nitric oxide + water

(g) Ammonia + oxygen $\to$ nitric oxide + water

(h) Barium chloride + sulphuric acid $\to$ barium sulphate + hydrochloric acid

(i) Zinc sulphide + oxygen $\to$ zinc oxide + sulphur dioxide

(j) Aluminium carbide + water $\to$ aluminium hydroxide + methane

(k) Iron pyrites ( $F e S_{2}$ ) + oxygen $\to$ ferric oxide + sulphur dioxide

(l) Potassium permanganate + hydrochloric acid $\to$ potassium chloride + manganese chloride + chlorine + water

(m) Aluminium sulphate + sodium hydroxide $\to$ sodium sulphate + sodium meta aluminate + water.

(n) Aluminium + sodium hydroxide + water $\to$ sodium meta aluminate + hydrogen

(o) Potassium dichromate + sulphuric acid $\to$ potassium chloride + chromium chloride + water + chlorine.

(p) Sulphur + nitric acid $\to$ sulphuric acid + nitrogen dioxide + water.

(q) Sodium chloride + manganese dioxide + sulphuric acid $\to$ sodium hydrogen sulphate + manganese sulphate + water + chlorine.

Q. (a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C.
(b) In the Contact process for the manufacture of the sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead, a two-steps procedure is used. Write the equations for the two steps involved in D.
(c) What type of substance will liberate sulphur dioxide from sodium sulphite to step E?
(d) Write the equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F.
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The solution which liberates sulphur dioxide gas, from sodium sulphite :

The correct option is C Dilute sulphuric acidSodium sulphite reacts with dilute sulphuric acid in the manner shown below:Na2SO3(s)+H2SO4(l)→Na2SO4+SO2(g)+H2O So, SO2 is liberated as a product in the above reaction.

The correct option is C Dilute sulphuric acid
Sodium sulphite reacts with dilute sulphuric acid in the manner shown below:
$N a_{2} S O_{3} (s) + H_{2} S O_{4} (l) \to N a_{2} S O_{4} + S O_{2} (g) + H_{2} O$
So, $S O_{2}$ is liberated as a product in the above reaction.