Answer:
\[\underset{\underset{65.3g}{\mathop{1\,mol}}\,}{\mathop{Zn}}\,+2HCl\to
ZnC{{l}_{2}}+\underset{\underset{22.7L\,at\,STP}{\mathop{1\,mol}}\,}{\mathop{{{H}_{2}}}}\,\]
Volume of \[{{H}_{2}}\] gas
liberated by 32.65 g Zinc a \[\frac{22.7}{65.3}\times 32.65\]
\[=11.35L\]at STP
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