2/

$\text{-Oxit axit :}$

$\text{+SO2 : lưu huỳnh đioxit}$

$\text{+CO2 : cacbon đioxit}$

$\text{+P2O5 : điphotpho pentaoxit}$

$\text{-Oxit bazo :}$

$\text{+Na2O : natri oxit}$

$\text{+MgO : magie oxit}$

$\text{+Fe2O3 : sắt III oxit}$

$3/$

$a/2KMnO4 \overset{t^o}\to K_2MnO_4+MnO2+O2$

$b/n_{O_2}=0,672/22,4=0,03mol$

Theo pt :

$n_{KMnO_4}=2. n_{O_2}=2.0,03=0,06mol$

$⇒m_{KMnO_4}=0,06.158=9,48g$

$c/n_{KMnO_4}=15,8/158=0,1mol$

Theo pt :

$n_{O_2}=1/2.n_{KMnO_4}=1/2.0,1=0,05mol$

$⇒V_{O_2}=0,05.22,4=1,12l$

Bài 2:

-Oxit bazo: 

+$Na_2O$: Natri oxit

+$MgO$: Magie oxit

+$Fe_2O_3$ : Sắt (III) oxit

-Oxit axit:

+$SO_2$: Lưu huỳnh đioxit

+$CO_2$: Cacbon đioxit

+$P_2O_5$: Điphotpho Pentaoxit

Bài 3:

a) $2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$

b) $n_{O_2}=\dfrac{0,672}{22,4}=0,03\,(mol)$

$\to n_{KMnO_4}=2n_{O_2}=0,06\,(mol)\\\to m_{KMnO_4}=0,06\times158=9,48\,(g)$

c) $n_{KMnO_4}=\dfrac{15,8}{158}=0,1\,(mol)\\\to n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,05\,(mol)\\\to V_{O_2}=0,05\times22,4=1,12\,(l)$