$1/$

$n_{Al}=2,16/27=0,08mol$

$4Al  +  3O_2\overset{t^o}{\longrightarrow}2Al_2O_3$

$0,08→0,06$

$⇒V_{O_2}=0,06.22,4=1,344l$

$2/$

$n_{Na}=2,3/23=0,1mol$

$4Na+O_2\overset{t^o}{\longrightarrow}2Na_2O$

$0,1→0,025$

$⇒V_{O_2}=0,025.22,4=0,56l$

$3/$

$n_{Mg}=4,32/24=0,18mol$

$2Mg+O_2\overset{t^o}{\longrightarrow}2MgO$

$0,18→0,09$

$⇒V_{O_2}=0,09.22,4=2,016l$

 

4Al+3O2-> 2AL2O3

nAl=2,16/27=0,08 mol

n02=(0,08*3)/2=0,12mol

VO2= n.22,4=0,12*22,4=2,688 lít

B2: 4Na+O2->2Na2O

nNa= 2,3/23=0,1mol

nO2= 0,025mol

Vo2= 0,025*22,4= 0,56 lít

B3 2Mg+O2-> 2MgO

nMg= 4,32/24=0,18 mol

nO2=0,18 mol

Vo2= 0,18*22,4=4,032