Question

4Al(s) + 3O2(g) → 2Al2O3(s) starting with 6.0 mol of Al, how many moles of O2 must be provided, at a minimum, to complete the reaction?

a. 9.0 mol O2
b. 6.0 mol O2
c. 4.5 mol O2
d. 3.0 mol O2

Answer 1

A minimum of 4.5 moles of O2 is required to react with 6.0 moles of Al according to the stoichiometry of the chemical reaction 4Al(s) + 3O2(g) → 2Al2O3(s).

Step-by-step explanation:

The question pertains to the stoichiometry of a chemical reaction involving aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃). According to the balanced equation, 4Al(s) + 3O₂(g) → 2Al₂O₃(s), 4 moles of Al react with 3 moles of O₂. If we begin with 6.0 moles of Al, we can find the amount of O₂ needed by setting up a proportion based on the mole ratio from the equation.

Using the ratio of Al to O₂ from the equation 4:3, we can perform a simple calculation:

(6.0 moles Al) * (3 moles O₂) / (4 moles Al) = 4.5 moles O₂

Therefore, the minimum amount of O₂ needed to react with 6.0 moles of Al is 4.5 moles of O₂.