Question

Answer 1

nAl = 5,4/27 = 0,2 (mol)

PTHH: 4Al + 3O2 -> (t°) 2Al2O3

Mol: 0,2 ---> 0,15

VO2 = 0,15 . 22,4 = 3,36 (l)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 0,15 . 2 = 0,3 (mol)

mKMnO4 = 0,3 . 158 = 47,4 (g)

Answer 2

4Al+3O₂-to>2Al₂O₃

0,2---0,15

2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,3---------------------------------------0,15

n _Al=\(\dfrac{5,4}{27}\)=0,2 mol

=>V_O2=0,15.22,4=3,36l

b)m _KMnO4=0,3.158=47,4g