Question

The amount of metallic $Zn$ (Atomic weight $=65.4$) required to react with aqueous sodium hydroxide to produce $1$g of $H_2$ is:

• A. $32.7$g
• B. $98.1$g
• C. $65.4$g
• D. $16.3$g

Answer 1

Answer: (A)

$32.7$g

$Zn+2NaOH\to N{a}_{2}Zn{O}_2+H_2$

From the above balenced equation it is clear that $1$ mole of $Zn$ that is $65.4gr$ of $Zn$ reacts with sodium hydroxide to give $1$ mole of $H_2$ that is $2gm$

So to produce $1gr$ of $H_2$ the $Zn$ required is:

$=\frac{65.4}{2}$

$=32.7g$

Hence option A is correct.