\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,3}{1}>\dfrac{0,5}{2}\Rightarrow HCl.hết,Zn.dư\\ Chất.sau.phản.ứng:ZnCl_2,H_2,Zn\left(dư\right)\\ n_{ZnCl_2}=n_{H_2}=n_{Zn\left(p.ứ\right)}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ n_{Zn\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\\ m_{ZnCl_2}=136.0,25=34\left(g\right)\\ m_{H_2}=0,25.2=0,5\left(g\right)\\ m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\)
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