Question #7396d

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Answer 1 · 2016-06-23T10:34:48

Option=1

$ZnCl_2" dissolves in excess NaOH"$
$"as well as in excess "NH_3$

In aqueous solution the reactions occuring can be represented by following equations.

For $NaOH$

$ZnCl_2(aq)+2NaOH(aq)=Zn(OH)_2darr+2NaCl(aq)$

$Zn(OH)_2+2NaOH(aq)=Na_2ZnO_2("soluble")+2H_2O$

For $NH_3(aq)=NH_4OH$

$ZnCl_2(aq)+2NH_4OH(aq)=Zn(OH)_2darr+2NH_4Cl(aq)$

$Zn(OH)_2+4NH_4OH(aq)=Zn(NH_3)_4(OH)_2("soluble")+4H_2O$