Question

How many grams of oxygen is needed to completely react with 9.30 moles of aluminum

Answer 1

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