Question

Balance this reaction: Al(s) + O2(g) → Al2O3(s)

How many moles of oxygen will be needed to react with aluminum metal to produce 12.0 moles of aluminum oxide?

6. Using the reaction listed in question 5 (above), how many moles of aluminum oxide will be produced by 40.0 moles of aluminum reacting completely with a boat load (that is a lot) of oxygen?

Answer 1

Hello!

A) The balanced reaction is the following:

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

To determine the required moles of Oxygen, we can use the following conversion factor to go from moles of Aluminum Oxide to moles of Oxygen. The coefficients from the equation will be used for this conversion factor:


`12 moles Al_2O_3* (3 moles O_2)/(2 moles Al_2O_3)= 18 moles O_2`

So, 18 moles of Oxygen are required to form 12 moles of Aluminum Oxide.

2) This question is similar to the above, the only difference is the mole ratio that will be used. In this case, the mole ratio would be Al₂O₃/Al because Aluminum will be the limiting reagent (in the lowest amount):


`40 moles Al * (2 moles Al_2O_3)/(4 moles Al)=20 moles Al_2O_3`

So, 20 moles of Al₂O₃ would be produced by the complete reaction of 40 moles of Al

Have a nice day!