Theo gt ta có: $n_{Al}=0,3(mol)$
$2Al+6HCl\rightarrow 2AlCl_3+3H_2$
Ta có: $n_{AlCl_3}=0,3(mol)\Rightarrow m=40,05(g)$
$n_{H_2}=1,5.n_{Al}=0,45(mol)\Rightarrow V_{H_2}=10,08(l)$
$n_{HCl}=0,9(mol)\Rightarrow C_M=1,8$
nAl = 8.1/27 = 0.3 (mol)
2Al + 3H2SO4 => Al2(SO4)3 + 3H2
0.3_____0.45______0.15_____0.45
CM H2SO4 = 0.45/0.5 = 0.9 (M)
mAl2(SO4)3 = 0.15*342 = 51.3(g)
VH2 = 0.45*22.4 = 10.08 (l)