ZnCl2-NaHCO3 AB-C-H2O B-NaOH - DIdentify the compound D present in the solution-

The correct option is D Na_2ZnO_2 ZnCl_2+2NaHCO_3→ Zn(HCO_3)_2+2NaCl Zn(HCO_3)_2→ ZnCO_3+H_2O+CO_2 ZnCO_3+2NaOH→ Zn(OH)_2+Na_2CO_3 Zn( ...

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Byju's Answer

Standard X

Chemistry

Concentration of Ore

ZnCl2+NaHCO3 ...

Question

$Z n C l_{2} + N a H C O_{3} H e a t - -- \to (A) H e a t - -- \to (B) + (C) ↑ + H_{2} O (B) + N a O H \to D$
Identify the compound $D$ present in the solution:

Z n C O_{3}

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Z n (O H)_{2}

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Z n O

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N a_{2} Z n O_{2}

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Solution

The correct option is D $N a_{2} Z n O_{2}$
$Z n C l_{2} + 2 N a H C O_{3} \to Z n (H C O_{3})_{2} + 2 N a C l$

$Z n (H C O_{3})_{2} \to Z n C O_{3} + H_{2} O + C O_{2}$

$Z n C O_{3} + 2 N a O H \to Z n (O H)_{2} + N a_{2} C O_{3}$

$Z n (O H)_{2} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H_{2} O$
Option D is correct.

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Similar questions

Q. Colourless solution of (A) gives white precipitate (B) with

A g N O_{3}

solution, soluble in aq.

N H_{3}

. (A) also gives white precipiatete (C) with NaOH soluble in excess of it forming (D). (D) gives white precipiate (E) with

H_{2} S

.(A), (B), (C), (D) and (E) are as follows.

(A) :

Z n C l_{2}

(B) :

A g C l

Z n (O H)_{2}

(D) :

N a_{2} Z n O_{2}

(E) :

Z n S

Q. In the following reactions,

Z n O

is respectively acting as a/an:
A)

Z n O + N a_{2} O \to N a_{2} Z n O_{2}

Z n O + C O_{2} \to Z n C O_{3}

Q. In the following reactions,

Z n O

is respectively acting as a/an:
(a)

Z n O + {N a}_{2} O ⟶ {N a}_{2} Z n O_{2}

(b)

Z n O + C O_{2} ⟶ Z n C O_{3}

Q. A white amorphous powder (A) when heated gives a colourless gas (B), which turns lime water milky and the residue (C) which is yellow when hot but white when cold. The residue (C) dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dilute HCl with the evolution of a gas which is identical in all respects with (B). The solution of (A) as obtained above gives a white precipitate (D) on addition of excess of

N H_{4} O H

and on passing

H_{2} S

. Another portion of this solution gives initially a white precipitate (E) on addition of NaOH which dissolves in excess of it. Compounds (A) to (D) are identified as:
(A) :

Z n C O_{3}

, (B) :

C O_{2}

, (C) :

Z n O

, (D) :

Z n S

and (E) :

Z n (O H)_{2}

If true enter 1, else enter 0.

Q. Assertion :

Z n {(O H)}_{2}

is dissolved in both

{N H}_{4} O H

and

N a O H

solution. Reason: Both

N a O H

and

{N H}_{4} O H

being basic can dissolve amphoteric

Z n {(O H)}_{2}

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ZnCl2+NaHCO3Heat−−−→(A)Heat−−−→(B)+(C)↑+H2O(B)+NaOH→DIdentify the compound D present in the solution:

The correct option is D Na2ZnO2ZnCl2+2NaHCO3→Zn(HCO3)2+2NaClZn(HCO3)2→ZnCO3+H2O+CO2ZnCO3+2NaOH→Zn(OH)2+Na2CO3Zn(OH)2+2NaOH→Na2ZnO2+2H2OOption D is correct.

$Z n C l_{2} + N a H C O_{3} H e a t - -- \to (A) H e a t - -- \to (B) + (C) ↑ + H_{2} O (B) + N a O H \to D$
Identify the compound $D$ present in the solution: