\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(m_{ddHCl}=\dfrac{7,3.100}{7,3\%}=100\left(g\right)\)
mH2 = 0,1 . 2 = 0,2(g)
mZnCl2 = 0,1 . 136 = 13,6(g)
mdd = mZn +mddHCl - mH2 = 6,5 + 100 - 0,2 = 106,3(g)
=> \(C\%_{ZnCl_2}=\dfrac{13,6.100}{106,3}=12,8\%\)