Question

If 67g of aluminum and 58g of oxygen react to completion, which reagent is the limiting reagent?

Also, calculate the greatest number of moles of aluminum oxide that can be formed under these conditions

Answer 1

The balanced equation of Al and O2 is as follows : -

4Al + 3O2 = 2 Al2O3

Now calculate the moles of Al and O2 as follows:

Numbe r of moles = amount in g/ molar mas

Al moles = 67 g/ 26.98 g/ moles

= 2.48 moles Al

O2 moles = 58 g / 31.99880 g/ moles

= 1.81 moles O2

In this reaction the limiting agent is O2 . The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Now calculate moles of aluminum oxide as follows:

1.81 moles O2 * 2 mole Al2O3/ 3 moles O2

= 1.21 moles Al2O3