\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
=> nZnCl2 = 0,2 (mol)
=> mZnCl2 = 0,2.136 = 27,2 (g)
=> B
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+Cl_2\underrightarrow{t^o}ZnCl_2\\ Mol:0,2\leftarrow0,2\leftarrow0,2\\ m_{ZnCl_2}=136.0,2=27,2\left(g\right)\)