Question

The amount of zinc required to produce $1.12\text{mL}$ of $H_2$ at STP on treatment with $dil.HCl$ will be: (Given: Molar mass of $Zn$= $65g/mol$)

• A. $60\text{g}$
• B. $0.65\text{g}$
• C. $32.5\times {10}^{-4}\text{g}$
• D. $6.5\text{g}$

Answer 1

The correct option is C $32.5\times {10}^{-4}\text{g}$

The reaction of Zinc with $dil.HCl$ gives:
$Zn+2HCl\to ZnC{l}_2+H_2$
$22400\text{mL}$ of $H_2$ is produced from $65\text{g}$ of zinc.
So, $1.12\text{mL}$ of $H_2$ will be produced from $(\frac{65}{22400}\times 1.12)=32.5\times {10}^{-4}\text{g}$ of zinc.