$a,PTPƯ:4Al+3O_2\xrightarrow{t^o} 2Al_2O_3$
$n_{Al}=\dfrac{4,5}{27}=\dfrac{1}{6}mol.$
$Theo$ $pt:$ $n_{O_2}=\dfrac{3}{4}n_{Al}=0,125mol.$
$⇒V_{O_2}=0,125.22,4=2,8l.$
$b,PTPƯ:2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2$
$Theo$ $pt:$ $n_{KMnO_4}=2n_{O_2}=0,25mol.$
$⇒m_{KMnO_4}=0,25.158=39,5g.$
chúc bạn học tốt!