\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,125 0,125 0,125
\(V_{H_2}=0,125.224=2,8\left(l\right)\\
m_{\text{dd}}=8,125+50-0,125.2=57,875\left(g\right)\\
C\%_{ZnCl_2}=\dfrac{0,125.136}{57,875}.100\%=29,374\%\)