$n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol \\PTHH : \\Al_2O_3+6HCl\to 2AlCl_3+3H_2O \\a.Theo\ pt : \\n_{HCl}=6.n_{Al_2O_3}=6.0,1=0,6mol \\⇒m_{HCl}=0,6.36,5=21,9g \\⇒C\%_{HCl}=\dfrac{21,9}{200}.100\%=10,95\% \\b.Theo\ pt : \\n_{AlCl_3}=2.n_{Al_2O_3}=2.0,1=0,2mol \\⇒m_{AlCl_3}=0,2.133,5=26,7g \\m_{dd\ spu}=10,2+200=210,2g \\⇒C\%_{AlCl_3}=\dfrac{26,7}{210,2}.100\%=12,7\%$