$\rm PTHH:4Al+3O_2 \xrightarrow{t^o} 2Al_2O_3$

$\rm n_{O_2}=\dfrac{19,2}{32}=0,6(mol)$

$\rm n_{Al}=\dfrac{4}{3}n_{O_2}=0,8(mol)$

`=>`$\rm a=m_{Al}=0,8.27=21,6(g)$

$\rm n_{Al_2O_3}=\dfrac{2}{3} n_{O_2}=0,4(mol)$

`=>`$\rm b=m_{Al_2O_3}=0,4.102=40,8(g)$