$n_{Al}=81/27=3mol$
$n_{O_2}=3,36/22,4=0,15mol$
$ 4Al+ 3O_2\overset{t^o}{\longrightarrow}2Al_2O_3$
Theo pt : $4 mol$ $3 mol$
Theo đbài: $3 mol$ $0,15mol$
⇒Sau phản ứng Al dư
Theo pt :
$n_{Al pư}=4/3.n_{O_2}=4/3.0,15=0,2mol$
$⇒n_{Al dư}=3-0,2=2,8mol$
$⇒m_{Al dư}=2,8.27=75,6g$
$n_{Al_2O_3}=2/3.n_{O_2}=2/3.0,15=0,1mol$
$⇒m_{Al_2O_3}=0,1.102=10,2g$
$b/$
$2KMnO_4\overset{t^o}{\longrightarrow}K_2MnO_4+MnO_2+O_2$
Theo pt :
$n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3mol$
$⇒m_{KMnO_4}=0,3.158=47,4g$