Question

Aqueous zinc oxide reacts with excess of NaOH, the final products are:
a.) $Zn(OH)_{ 2 }$
b.) $Na_{ 2 }ZnO_{ 2 }$
c.) Both A and B
d.) None of them

Answer 1

Hint: In this question, you just need to focus on the word excess. Aqueous zinc oxide will dissolve in alkalis to give soluble compounds. Now give your answer accordingly.

Complete step by step answer:

Zinc hydroxide can be prepared by adding sodium hydroxide solution, but not in excess, to a solution of any zinc salt. A white precipitate will be seen:
$Zn^{ 2+ }\quad +\quad 2OH^{ - }\quad\rightarrow\quad Zn(OH)_{ 2 }$
If excess sodium hydroxide is added, the precipitate of zinc hydroxide will dissolve, forming a colorless solution of zincate ion:
$Zn(OH)_{ 2 }\quad +\quad 2OH^{ - }\quad \rightarrow\quad Zn(OH)_{ 4 }^{ 2- }$
Solutions of sodium zincate may be prepared by dissolving zinc, zinc hydroxide, or zinc oxide in an aqueous solution of sodium hydroxide.
Hence, we can say that zinc oxide reacts with an excess of sodium hydroxide to produce sodium zincate and water. This reaction takes place at a temperature of 500-${ 600 }^{ 0 }$C.
$ZnO\quad +\quad 2NaOH\quad \rightarrow\quad Na_{ 2 }ZnO_{ 2 }\quad +\quad H_{ 2 }O$

Therefore, we can conclude that the correct answer to this question is option B.

Note: Sodium zincate refers to anionic zinc oxides or hydroxides, depending on conditions. In the applications of these materials, the exact formula is not necessarily important and it is likely that aqueous zincate solutions consist of mixtures.
$Na_{ 2 }ZnO_{ 2 }$ is a type of oxo zincate.