$n_{Zn}=\dfrac{6,5}{65}=0,1mol \\m_{HCl}=200.14,6\%=29,2g \\⇒n_{HCl}=\dfrac{29,2}{36,5}=0,8mol \\PTHH :$
$Zn + 2HCl\to ZnCl_2+H_2↑$
Theo pt : 1 mol 2 mol
Theo đbài : 0,1 mol 0,8 mol
Tỉ lệ : $\dfrac{0,1}{1}<\dfrac{0,8}{2}$
⇒Sau phản ứng HCl dư
$Theo\ pt : \\n_{HCl\ pư}=2.n_{Zn}=2.0,1=0,2mol \\⇒n_{HCl\ dư}=0,8-0,2=0,6mol \\⇒m_{HCl\ dư}=0,6.36,5=21,95(g) \\n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1 mol \\⇒m_{ZnCl_2}=0,1.136=13,6(g) \\m_{dd\ spư}=6,5+200-0,1.2=206,3(g) \\⇒C\%_{HCl\ dư}=\dfrac{21,9}{206,3}.100\%=10,62\% \\C\%_{ZnCl_2}=\dfrac{13,6}{206,3}.100\%=6,59\%$