Balanced chemical equation:
* moles of oxygen
4 Al + 3 O2 = 2 Al2O3
4 moles Al -------------- 3 moles O2
9.30 moles Al ---------- moles O2
moles O2 = 9.30 * 3 / 4
moles O2 = 27.9 / 4 => 6.975 moles of O2
Therefore:
Molar mass O2 = 31.9988 g/mol
n = m / mm
6.975 = m / 31.9988
m = 6.975 * 31.9988
m = 223.19 of O2
Answer:
moles of calcium hydroxide= 21.75 mol
a) 43.5 mol
b) 7.25 mol
Explanation:
Please see the attached picture for the full solution.
First a balanced reaction equation must be established:

→

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols
Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[
4Al +
3 O₂ → 2 Al₂O₃]
∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
=
3.63 mol (to 2 sig. fig.)
Thus it can be concluded that
3.63 moles of oxygen is needed to react completely with 145 g of aluminum.