Final answer:
Removing aluminum oxide (Al2O3) from the reaction mixture will cause the reaction to shift to the right, increasing the consumption of aluminum (Al) and oxygen (O2) to produce more Al2O3, thereby decreasing the concentration of Al and O2.
Step-by-step explanation:
The effect of removing aluminum oxide (Al2O3) from the mixture in the reversible reaction 4Al + 3O2 ⇌ 2Al2O3 is predicted by Le Chatelier's principle. According to this principle, if a product is removed from a system at equilibrium, the equilibrium will shift to produce more of that product. Therefore, removing Al2O3 will shift the equilibrium to the right, resulting in an increase in the consumption of aluminum (Al) and oxygen (O2) gas to form more Al2O3. This shift will decrease the concentration of aluminum and oxygen in the reaction mixture, prompting the reaction to make more aluminum oxide.
The balanced thermochemical equation for the refining process in which aluminum oxide is turned into aluminum and oxygen is:
2 Al2O3(s) → 4 Al(s) + 3 O2(g)
This reaction involves the reduction of aluminum ions to aluminum at the cathode during electrolysis, and the generation of oxygen gas at the anode.