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Q.
The amount of Zinc (atomic weight $=65$ ) necessary to produce $224\, mL$ of $H _{2}$ by the reaction with an acid will be
Some Basic Concepts of Chemistry
Solution:
Atomic weight of $Zn =95$ and volume of $H _{2}=224\, mL$
$Zn + H _{2} SO _{4} \longrightarrow ZnSO _{4}+ H _{2}$
$\frac{\text { Volume of } H _{2}}{22400}=\frac{\text { Weight of } Zn }{\text { Atomic weight of } Zn }$
Wt of $Z n=\frac{224 \times 65}{22400}=0.65 \,g$