Question

Ce volum de aer(20% oxigen in procente de volum) este necesar pt. arderea totala a 5,03 g Aluminiu ?
Ce produs rezulta si in ce cantitate ?

Answer 1

MAl=27g/mol

n=m/M=5,03/27=0,186 moli Al

4Al + 3O2 => 2Al2O3

4*27g Al.........3 moli O2
5,03g Al.........x=0,13972 moli O2

4 moli Al.........2 moli Al2O3
0,186 moli Al.......x moli Al2O3

x=0,0931 moli Al2O3

m Al2O3=9,5g


VO2=0,13972*22,4=3,1297 L

Vaer(20℅O2) = 5*VO2=5*3,1297=15,648 L