MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Preparations and Uses of Caustic Soda

Element A whi...

Question

Element A which belongs to group 12 and period 4 is a bluish white metal. A reacts with conc. $H_{2} S O_{4}$ and gives B with the liberation of $S O_{2}$ . With $N a O H$ . A gives compound C. Identify A, B and C. Explain the reaction.

Open in App

Solution

$A \to Z i n c (Z n)$
$B \to Z i n c s u l p h a t e (Z n S O_{4})$
$C \to S o d i u m Z i n c a t e (N a_{2} Z n O_{2})$
Reaction:
Zinc reacts with hot con. $H_{2} S O_{4}$ formating $Z n S O_{4}$
$Z n (A) + 2 H_{2} S O_{4} \to Z n S O_{4} (B) + S O_{2} ↑ + 2 H_{2} O$
Zinc dissolves in hot NaOH solution formating soluble zincate ion.
$Z n + 2 N a O H \to N a_{2} Z n O_{2} S o d i u m Z i n c a t e (C) + H_{2}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. An element (A) belongs to group number

12

and period number

4

, bluish white in colour reacts with Conc.

H_{2} S O_{4}

to give its salt (B) with the liberation of

S O_{2}

gas. Compound (B) reacts with sodium bicarbonate to give (C) which is used as ointment for curing skin diseases. Identify A, B and C and explain the reactions.

Q. (a) Compound

(A)

with molecular formula

C_{6} H_{6} O

gives violet colour with neutral

F e {C l}_{3}

(A)

C H {C l}_{3}

N a O H

gives two isomers

(B)

(C)

with molecular formula

C_{7} H_{6} O_{2}

(A)

reacts with ammonia at

473 K

in the presence of

Z n {C l}_{2}

(D)

with molecular formula

C_{6} H_{7} N

(D)

undergoes carbylamine test. Identify

(A), (B), (C)

(D)

and explain the reactions.
(b)

(A)

is reddish brown metal. It belongs to group 11 and period 4 of the periodic table. When heated below

1370 K

(A)

given a black compound

(B)

1370 K

(A)

gives a red compound

(C)

. With concentrated nitric acid,

(A)

{N O}_{2}

gas and gives compound

(D)

(A), (B), (C)

(D)

. Explain the reactions.

Q. An element (A) belongs to period number 4 (four) and group number 11 (eleven) and is extracted from its pyrite ore. Element (A) reacts with oxygen at two different temperatures forming compounds (B) and (C). Element (A) also reacts with conc.

H N O_{3}

to give compound (D) with the evolution of

N O_{2}

. Identify (A), (B), (C) and (D). Explain the reactions.

Q. An organic compound (A) of molecular formula

C_{6} H_{6} O

gives violet colour with neutral ferric chloride. Compound (A) when heated with zinc dust gives a hydrocarbon (B). Also compound (A) reacts with phthalic anhydride and conc.

H_{2} S O_{4}

to give compound (C) of molecular formula

C_{20} H_{14} O_{4}

. Identify (A), (B) and (C). Explain the reactions.

Q. a) An organic compound (A) of molecular formula

C_{2} H_{6} O

liberates hydrogen with metallic sodium. Compound (A) on heating with conc.

H_{2} S O_{4}

at 440K gives an alkene (B). Compound (B) when oxidised by baeyer's reagent gives compound (C) of molecular formula

C_{2} H_{6} O_{2}

. Identify A,B and C and explain the above reactions.
b) The metal (A) is extracted from its sulphide ore. On treatment with dilute nitric acid metal (A) gives a compound (B), which is also known as lunar caustic. (B) on heating at 723K gives (C) and

O_{2}

. Identify A,B and C and explain the reactions.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Sodium Hydroxide

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Preparations and Uses of Caustic Soda

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon