2KMnO4⇒K2MnO4+MnO2+O2

⇒chất rắn A:K2MnO4, MnO2

nKMnO4=$\frac{15.8}{158}$=0.1(mol)

1.

nMnO2=nK2MnO4= $\frac{1}{2}$ nKMnO4=0.05

mA=mMnO2+mK2MnO4=(0.05×87)+(0.05+197)=14.2(g)

2.

nO2=nMnO2=0.05

⇒VO2=0.05×22.4=1.12(l)

3.

4Al+3O2⇒2Al2O3

⇒chất rắn B:Al2O3

nAl= $\frac{5.4}{27}$ =0.2(mol)

Mà nAl2O3= $\frac{2}{4}$ nAl=0.1(mol)

⇒mB=0.1×(27×2+16×3)=10.2

$2KMnO4→K2MnO4+MnO2+O2$

$1/$

Chất rắn $: K2MnO4; MnO2 $

theo pt :

$nK2MnO4=nMnO4=1/2.nKMnO4=0,05mol$

$⇒m_{A}=m_{K_{2}MnO_{4}}+m_{MnO_{2}}=197.0,05+87.0,05=14,2g$

$2/$

theo pt :

$nO_{2}=1/2.n_{KMnO4}=0,05mol$

$⇒V_{O_{2}}=0,05.22,4=1,12l$

$3/$

$4Al+3O2→2Al2O3$

Chất rắn$ :Al2O3$

theo pt :

$nAl=5,4/26=0,2mol$

$⇒nAl2O3=1/2.nAl=1/2.0,2=0,1mol$

$⇒mAl2O3=0,1.102=10,2g$