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Q. The amount of zinc (at. wt. = 65) necessary to produce $224\, ml$ of $H_2$ by the reaction with an acid will be

AIIMSAIIMS 1995

Solution:

$Zn + H_{2}SO_{4} \to ZnSO_{4} + H_{2}$

$65 \,g$ Zn will give 22.4 litre of $H_{2}$ at STP

$\Rightarrow \, 224 \,ml $ of $H_{2}$ corresponds to $\frac{65}{22400}\times224$

$= 0.65 \,g \,Zn$