Answer:
AlF₃ is the excess reactant
Explanation:
We are given;
- 13 moles of AlF₃
- 9 moles of O₂
We are required to identify the excess reactant;
First; we write a balanced equation for the reaction between O₂ and AlF₃
- O₂ and AlF₃ reacts to form Al₂O₃ and F₂
- Therefore; the equation of the reaction is
4AlF₃ + 3O₂ → 2Al₂O₃ + 6F₂
Second, determine the mole ratio of AlF₃ to O₂
- From the equation 4 moles of AlF₃ reacts with 3 moles of O₂
- Therefore; the mole ratio of AlF₃ to O₂ is 4 : 3
Third; we determine the excess reactant.
- From the mole ration, 13 moles of AlF₃ will require 9.75 moles (13×3/4) of O₂ but only 9 moles of oxygen are available.
- On the other hand, 9 moles of oxygen would require 12 moles (9×4/3) AlF₃. But we have 13 moles of AlF₃, which means AlF₃ is in excess.
Therefore; Oxygen gas is the rate limiting reactant and AlF₃ is the excess reactant (1 mole excess).
