$n_{O_2}=6,72/22,4=0,3mol$
$a/$
$4Al+3O_2\overset{t^o}\to 2Al_2O_3$
$b/$
Theo pt :
$n_{Al}=4/3.n_{O_2}=4/3.0,3=0,4mol$
$⇒m_{Al}=0,4.27=10,8g$
$c/$
$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$
Theo pt :
$n_{KMnO_4}=2.n_{O_2}=2.0,3=0,6mol$
$⇒m_{KMnO_4}=0,6.158=94,8g$