What is the maximum number of moles of Al2O3 that can be produced by the reaction of 0.40 mol of Al — Stoichiometry Chemistry Question

STEPS:

1. Write and Balance the Chemical Equation: The first step is to establish the balanced chemical equation for the synthesis of aluminum oxide from its elements: $4 Al(s) + 3 O_2(g) \rightarrow 2 Al_2O_3(s)$. This shows the stoichiometric relationship: 4 moles of $Al$ react with 3 moles of $O_2$ to produce 2 moles of $Al_2O_3$.
2. Identify the Limiting Reactant: A "limiting reactant" is the substance that is completely consumed first, stopping the reaction. To find it, divide the available moles of each reactant by its respective coefficient from the balanced equation:
* For $Al$: $0.40\text{ mol} / 4 = \mathbf{0.10}$
* For $O_2$: $0.40\text{ mol} / 3 \approx \mathbf{0.133}$
Since $0.10$ is less than $0.133$, aluminum ($Al$) is the limiting reactant; the reaction will stop once all $0.40$ moles of $Al$ are used.
3. Calculate Product Yield based on the Limiting Reactant: Use the stoichiometric ratio between the limiting reactant ($Al$) and the product ($Al_2O_3$) from the balanced equation ($4:2$ or $2:1$).
$$\text{Moles of } Al_2O_3 = 0.40\text{ mol Al} \times \left(\frac{2\text{ mol } Al_2O_3}{4\text{ mol Al}}\right) = \mathbf{0.20\text{ mol } Al_2O_3}$$
4. Conclusion: The maximum amount of $Al_2O_3$ that can be produced is 0.20 mol, which corresponds to option B.

WHY_OTHERS_WRONG:

• A) 0.10 mol: This value might be reached if a student correctly determined the "limiting factor" (0.10) but forgot to multiply by the coefficient of the product (2).
• C) 0.27 mol: This is the amount of $Al_2O_3$ that would be produced if oxygen were the limiting reactant ($0.40\text{ mol } O_2 \times 2/3 \approx 0.27$). While there is enough oxygen to make this much, there is not enough aluminum to support it.
• D) 0.33 mol: This distractor likely results from using an incorrect stoichiometric ratio (such as 3:1) or a computational error when comparing the reactants.
• E) 0.40 mol: This assumes a 1:1 molar ratio between the reactants and the product, which ignores the subscripts and coefficients required to balance the chemical equation for $Al_2O_3$.