\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{1}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
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