To determine the number of moles of oxygen required to produce 14 moles of Al2O3, we need to examine the balanced chemical equation for the reaction.
The balanced equation for the formation of Al2O3 from aluminum (Al) and oxygen (O2) is:
4 Al + 3 O2 → 2 Al2O3
From the balanced equation, we can see that 3 moles of oxygen (O2) are required to produce 2 moles of Al2O3.
Using this ratio, we can set up a proportion to find the number of moles of oxygen required:
3 moles O2 / 2 moles Al2O3 = x moles O2 / 14 moles Al2O3
Cross-multiplying the proportion, we get:
3 moles O2 * 14 moles Al2O3 = 2 moles Al2O3 * x moles O2
42 moles O2 = 2x
Dividing both sides by 2, we find:
x = 42 moles O2 / 2 = 21 moles O2
Therefore, to produce 14 moles of Al2O3, we would need 21 moles of oxygen.