$a,\\ 2M+3Cl_2\xrightarrow{t^0} 2MCl_3\\ n_{M}=n_{MCl_3}\\ \Rightarrow \dfrac{13,5}{M}=\dfrac{66,75}{M+35,5.3}\\ \Rightarrow M=27\\ \to Al\ (nhôm)\\ b,\\ n_{Al}=\dfrac{13,5}{27}=0,5(mol)\\ n_{Cl_2}=\dfrac{3}{2}.n_{Al}=0,75(mol)\\ n_{MnO_2}=n_{Cl_2}=0,75(mol)\\ m_{MnO_2}=0,75.87=65,25(g)\\ n_{HCl}=4.n_{Cl_2}=3(mol)\\ m_{dd\ HCl}=\dfrac{3.36,5}{37\%}=295,95(g)\\ V_{dd\ HCl}=\dfrac{295,55}{1,19}=248,7(ml)$