Question

Find the limiting reactant for each initial amount of reactants in the reaction 4Al(s) + 3O₂(g) → 2Al₂O₃(s)?

Answer 1

To find the limiting reactant in the reaction of Al with O₂, convert the given masses to moles, use stoichiometry to find how much Al₂O₃ each could produce, and the one that produces less is the limiting reactant. The leftover mass of the other reactant can be calculated from the difference between the initial mass and the mass that reacted.

Step-by-step explanation:

In the reaction 4Al(s) + 3O₂(g) → 2Al₂O₃(s), to find the limiting reactant when starting with 50.0 g of Al and 50.0 g of O₂, we need to perform mole-mass calculations for each reactant. We convert the masses of Al and O₂ to moles and then apply the stoichiometry of the balanced equation to determine how much product can be formed from each reactant. The reactant that produces the lesser amount of product is the limiting reactant, and any remaining amount of the other reactant is considered in excess.

Steps to determine the limiting reactant:

1. Convert the mass of Al and O₂ to moles using their molar masses.
2. Apply the stoichiometry from the balanced equation to calculate the amount of Al₂O₃ that could be formed from each reactant.
3. The reactant that can yield the smaller amount of Al₂O₃ is the limiting reactant.

Using this method, we find that the reactant resulting in less Al₂O₃ is the limiting reactant, and the mass of the leftover reactant can be calculated by determining how much of it was actually used in the reaction based on the amount of limiting reactant present.