\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.15................................0.15\)
\(m_{Zn}=0.15\cdot65=9.75\left(g\right)\)
\(Zn + 2HCl \to ZnCl_2 + H_2\\ n_{Zn} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ m_{Zn} = 0,15.65 = 9,75(gam)\)