1/
$nKMnO4=15,8/158=0,1mol$
$pthh:$
$2KMnO4→KMnO4+MnO2+O2$
$a/$
theo pt:
$nO2=1/2.nKMnO4=1/2.0,1=0,05mol$
$⇒V_{O2}=0,05.22,4=1,12l$
$b/$
$n_{Cu}=4,8/64=0,075mol$
$O2+2Cu→2CuO$
theo pt :
$n_{CuO}=2.n_{O_{2}}=2.0,075=0,15mol$
$⇒m_{CuO}=0,15.80=12g$
$2/$
gọi $CTHH$ là $Cr_{x}O_{y}$
theo quy tác hóa trị :
$x.III=y.II$
$⇒\frac{x}{y}=\frac{II}{III}=\frac{2}{3}$
$⇒x=2 ; y=3$
Vậy CTHH là $Cr_{2}O_{3}$