\(\begin{array}{l}
n_{Cu}=\frac{19,2}{64}=0,3(mol)\\
n_{Cl_2}=\frac{7}{22,4}=0,3125(mol)\\
Cu+Cl_2\xrightarrow{t^o}CuCl_2\\
n_{Cu}<n_{Cl_2}\to \rm Hiệu\,suất\,tính\,theo\,Cu\\
n_{CuCl_2(lt)}=n_{Cu}=0,3(mol)\\
\to H=\frac{28,35}{0,3.135}.100\%=70\%
\end{array}\)