2Al₂O₃ = 4Al + 3O₂
M(A₂O₃)=101.96 g/mol
m(Al₂O₃)=250 g
n(O₂)=3m(Al₂O₃)/{2M(Al₂O₃)}
n(O₂)=3*250/{2*101.96}=3.678 mol
The answer is 2.37 x10^22 atoms
It is hard to answer this because not much information is given.
Yeah it depends on what mixture
You never told us how old you are so how are we supposed to answer
