\(n_{Al}=\dfrac{9,45}{27}=0,35\left(mol\right)\\ PTHH:2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\\ \left(mol\right)...0,35\rightarrow........0,35\\ PTHH:AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ \left(mol\right)....0,35\rightarrow..........................1,05\\ m_{AgCl}=1,05.143,5=150,675\left(g\right)\)