$PTHH : \\Zn+2HCl\to ZnCl_2+H_2↑(1) \\ZnO+2HCl\to ZnCl_2+H_2O(2) \\n_{H_2}=\dfrac{4,48}{22,4}=0,2mol \\Theo\ pt\ (1) : \\n_{Zn}=n_{H_2}=0,2mol \\⇒m_{Zn}=0,2.65=13g \\m_{ZnO}=21,1-13=8,1g \\⇒n_{ZnO}=\dfrac{8,1}{81}=0,1mol \\Theo\ pt\ (1) : \\n_{ZnCl_2(1)}=n_{H_2}=0,2mol \\⇒m_{ZnCl_2(1)}=0,2.136=27,2g \\Theo\ pt\ (2) : \\n_{ZnCl_2(2)}=n_{ZnO}=0,1mol \\⇒m_{ZnCl_2(2)}=0,1.136=13,6g \\⇒∑m_{ZnCl_2}=27,2+13,6=40,8g \\Theo\ pt : \\n_{HCl}=2.n_{Zn}+2.n_{ZnO}=2.0,2+2.0,1=0,6mol \\⇒m_{HCl}=0,6.36,5=21,9g \\⇒m_{dd\ HCl}=\dfrac{21,9}{10\%}=219g \\m_{dd\ spu}=21,1+219-0,2.2=239,7g \\⇒C\%_{ZnCl_2}=\dfrac{40,8}{239,77}.100\%=17,02\%$
