Question

Excess of sodium hydroxide reacts with zinc to form:

• A. $Zn{H}_2$
• B. $N{a}_{2}Zn{O}_2$
• C. $ZnO$
• D. $Zn(OH{)}_2$

Answer 1

Answer: (B)

$N{a}_{2}Zn{O}_2$

As $Zn$ is less reactive than $Na$

But when $Zn$ reacts with excess of $NaOH$ in aqueous conditions it gives $N{a}_{2}Zn{O}_2$.

$Zn+2NaOH⟶N{a}_{2}Zn{O}_2+H_2$