`m_(HCl)=\frac{200.14,6}{100}=29,2(g)`
`n_(HCl)=\frac{29,2}{36,5}=0,8(mol)`
`n_(Zn)=\frac{6,5}{65}=0,1(mol)`
`Zn+2HCl->ZnCl_2+H_2`
`0,1` `0,2` `0,1` `0,1`
`m_(dd)=6,5+200-0,1.2=206,3(g)`
`m_(HCl dư)=36,5.0,6=21,9(g)`
`m_(ZnCl_2)=0,1.135=13,5(g)`
`C%_(HCl)=\frac{21,9}{206,3}.100=10,62%`
`C%_(ZnCl_2)=\frac{13,5}{206,3}.100=6,544%`