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
Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Answer:
Explanation:
The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.
Its the second one. Add the exponents and you'll get 7
A is the answer I know bc I take biology
