a) \(m_{HCl}=300.7,3\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2<---0,6------->0,2------->0,3
=> mAl = 0,2.27 = 5,4 (g)
b) \(m_{ddspư}=300+5,4-0,3.2=304,8\left(g\right)\)
=> \(C\%_{AlCl_3}=\dfrac{0,2.133,5}{304,8}.100\%=8,76\%\)