$n_{Al}=81/27=3mol$

$n_{O_2}=3,36/22,4=0,15mol$

                 $ 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 :

$n_{KMnO_4}=2.n_{O_2}=2.0,15=0,3mol$

$⇒m_{KMnO_4}=0,3.158=47,4g$

Đáp án: Al dư, m=2,7g

            \({m_{KMn{O_4}}}\)=47,4g

 

Giải thích các bước giải:

\(\begin{array}{l}
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
{n_{Al}} = 3mol\\
{n_{{O_2}}} = 0,15mol\\
\dfrac{{{n_{Al}}}}{4} > \dfrac{{{n_{{O_2}}}}}{3}
\end{array}\)

Al dư

\(\begin{array}{l}
{n_{A{l_2}{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = 0,1mol \to m = 10,2g\\
{n_{Al dư }} = 0,3 - (\dfrac{4}{3} \times 0,15) = 0,1mol \to m = 2,7g
\end{array}\)

\(2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\({n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,3mol \to m = 47,4g\)