Câu 3:
a,
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$4Al+3O_2\buildrel{{t^o}}\over\to 2Al_2O_3$
$\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}=0,15(mol)$
$\to V_{O_2}=0,15.22,4=3,36l$
b,
$2KClO_3\buildrel{{MnO_2, t^o}}\over\longrightarrow 2KCl+3O_2$
$\Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,1(mol)$
$\to m_{KClO_3}=0,1.122,5=12,25g$