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Byju's Answer

Standard XII

Chemistry

Oxides of Sulphur

Yellow brittl...

Question

Yellow brittle solid (A) gives a colourless gas (B) on heating in air. B on heating in the
presence of oxygen gives gas C. C is dissolved in D to form E. E on dilution with water gives
D. B when dissolved in alkali 'F' gives salt G.

From the above information answer the following questions.

Write the balanced equations for B to G.

S O_{2} + 2 N a O H \to {N a}_{2} S O_{3} + H_{2} O

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2 S O_{2} + O_{2} \to 2 S O_{3}

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S O_{2} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}

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H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}

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Solution

The correct option is A $S O_{2} + 2 N a O H \to {N a}_{2} S O_{3} + H_{2} O$
$S (A) + O_{2} Δ - \to {S O}_{2} (B) {S O}_{2} + \frac{1}{2} O_{2} ⟶ {S O}_{3} (C) {S O}_{3} + H_{2} {S O}_{4} (D) ⟶ H_{2} S_{2} O_{7} (E) H_{2} S_{2} O_{7} (E) + H_{2} O ⟶ 2 H_{2} {S O}_{4} (D) {S O}_{2} (B) + 2 N a O H (F) ⟶ {N a}_{2} S_{2} O_{3} (G) + H_{2} O$

Therefore the required reaction is $S O_{2} + 2 N a O H \to N a_{2} S O_{3} + H_{2} O$

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Similar questions

A black colored (insoluble in

H_{2} O

) solid A does not dissolve in

H N O_{3}

. Aqua regia can dissolve A forming B. B gives yellow precipitate C with

N a O H

. B also gives an orange ppt. D with

K I

. D dissolves in excess of

K I

forming E. E gives brown ppt. with

N H_{4}^{+}

salt in presence of

N a O H

. A is precipitated if

H_{2} S

gas is passed into the solution of B in

d i l . H C l

The compound C is :

Q. Mg burns in air to form two compound (A) and (B). (A) is sparingly soluble in water and the resulting solution is weakly basic. (B) dissolve in water to form (C) and gas (D) with dilute

H N O_{3}

gives (E). E on heating gives gas (F) which behaves as an oxidising agent too. Gas F on reaction with phosphorus produce an important dehydrating agent (G). The compound G is:

Q. Colourless salt (A) give white ppt (B) with NaOH; ppt (B) dissolves in excess of NaOH to form (C). On passing

H_{2} S

gas into (C) white ppt (D) appears. (A) also gives white ppt (E) with

A g N O_{3}

solution. What is (A)?

Q. The compound

A

on heating gives a colourless gas and a residue that is dissolved in water to obtain

B

. Excess of

C O_{2}

is bubbled through an aqueous solution of

B, C

is formed which is recovered in the solid form. Solid

C

on gentle heating gives back

A

. The compound

A

is:

Q. (i) A black mineral (A) on heating in presence of air gives a gas (B).
(ii) The mineral (A) on reaction with dilute

H_{2} S O_{4}

gives a gas (C) and solution of a compound (D).
(iii) On passing gas (C) into an aqueous solution (B), a white turbidity is obtained.
(iv) The aqueous solution of compound (D) on reaction with potassium ferricyanide gives a blue compound (E).

Compounds (A) to (E) are identified as:

(A) :

F e S

, (B) :

S O_{2}

, (C) :

H_{2} S

, (D) :

F e S O_{4}

and (E) :

K F e^{I} I [F e^{I} I I (C N)_{6}]

If true enter 1, else enter 0.

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The correct option is A SO2+2NaOH→Na2SO3+H2OS(A)+O2Δ−→SO2(B)SO2+12O2⟶SO3(C)SO3+H2SO4(D)⟶H2S2O7(E)H2S2O7(E)+H2O⟶2H2SO4(D)SO2(B)+2NaOH(F)⟶Na2S2O3(G)+H2OTherefore the required reaction is SO2+2NaOH→Na2SO3+H2O