1/
$n_{KMnO4}=47,49/158=0,3mol$
$2KMnO4\overset{t^o}\to K2MnO4+MnO2+O2$
$\text{Theo pt :}$
$n_{O_2}=1/2.n_{KMnO4}=1/2.0,3=0,15mol$
$⇒V_{O_2}=0,15.22,4=3,36l$
2/
$PTPU :$
$2M+O2\overset{t^o}\to 2MO$
$\text{Theo pt :}$
$n_M=n_{MO}$
$⇒\dfrac{6,5}{M}=\dfrac{8,1}{M+16}$
$⇒M=65(Zn)$
$\text{Vậy kim loại là kẽm (Zn)}$