$a.PTHH : \\4Al+3O_2\overset{t^o}\to 2Al_2O_3(1) \\n_{Al}=\dfrac{5,4}{27}=0,2mol \\Theo\ pt\ (1) : \\n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.0,2=0,15mol \\⇒V=V_{O_2}=0,15.22,4=3,36l \\b.PTHH : \\2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2(2) \\Theo\ pt : \\n_{KMnO_4\ lt}=2.n_{O_2}=2.0,15=0,3mol$
Vì thực tế đã dùng dư $KMnO_4\ 10\%$
$⇒n_{KMnO_4\ tt}=0,3.110\%=0,33mol \\⇒m_{KMnO_4\ tt}=0,33.158=52,14g$