\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\)
\(n\left(Zn\right)=\dfrac{16,25}{65}=0,25\left(mol\right)\)
\(\left(1\right)\Rightarrow n\left(ZnCl_2\right)=n\left(Zn\right)=0,25\left(mol\right)\)
\(m\left(ZnCl_2\right)=0,5.\left(65+2.35,5\right)=0,25.136=34\left(g\right)\)