Answer:
Moles of Al2O3 formed = 2.7 mol (2 sig figs)
Step-by-step explanation:
In chemical equations, unless stated otherwise, the reactants and products will theoretically always remain in stoichiometric ratios.
The stoichiometry of a reaction is the relationship between the relative quantities of products and reactants, typically a ratio of whole integers.
Consider the following chemical reaction: aA + bB ⇒ cC + dD.
The stoichiometry of reactants to products in this reaction is the ratio of the coefficients of each species: a : b : c : d.
Now let's apply this knowledge to the question to be attempted:
first, we can start by writing out a balanced chemical equation, with states.
4Al(s) + 3O₂(g) ⇒ 2Al₂O₃(s). This is an example of a metal combustion.
Hence, the stoichiometry of this reaction is 4 : 3 : 2.
Next, we need to determine if, in the reaction, the substances ARE present in stoichiometric ratios. If they are not, then we need to identify the limiting reagent (the reactant which reacts completely), and the excess reagent (the reactant which is not completely used up). We can do this by inputting the mole values in the question into the ratios, until we figure out which doesn't match up.
Fast tracking, O₂ is the limiting reagent, and Al is the excess reagent.
Now we can use the limiting reagent, O₂ to calculate moles of Al₂O₃ produced. Moles of Al₂O₃ = moles of O₂ × 2/3 = 4×2/3 = 2.67 moles