Question

Given the equation $ 4Al\text{ }+\text{ }3{{O}_{2}}\text{ }\to \text{ }2A{{l}_{2}}{{O}_{3}} $ , which reactant is limiting if 0.32 mol Al and 0.26 mol $ {{O}_{2}} $ are available?

Answer 1

Hint: When a chemical reaction is complete, the limiting reagent is the material that has been entirely consumed. Because the reaction cannot proceed without it, the amount of product generated by the reaction is restricted by this reactant; typically, other reagents are present in excess of the amounts necessary to react with the limiting reagent. The precise quantity of reactant required to react with another element may be determined using stoichiometry. The limiting reagent will be completely consumed and the reaction will not reach stoichiometric completion if the reagents are not combined or present in these precise stoichiometric quantities.

Complete answer:
Limiting the amount of reagent In a reaction, the limiting reagent is the first to be entirely consumed, preventing any subsequent reactions from taking place. Because there is still some A in the products, reactant B is the limiting reagent in this reaction. As a result, when B was depleted, A was in excess.
Comparing the mole ratio of the amount of reactants utilised is one approach to find the limiting reagent. When there are just two reactants, this technique is most useful. The balanced chemical equation is used to calculate the amount of the other reactant (B) required to react with A.
If the actual amount of B present is greater than the necessary amount, B is in excess, and A is the limiting reagent. B is the limiting reagent if the amount of B present is less than what is necessary.
The reaction's stoichiometry requires 4 moles of aluminium for every 3 moles of diatomic oxygen. This indicates that the oxygen is the limiting reagent if the ratio of Al to diatomic oxygen is larger than $ \dfrac{4}{3} $ . Al is the limiting reagent if the ratio of Al to diatomic oxygen is less than $ \dfrac{4}{3} $ . According to the issue, the ratio of Al to diatomic oxygen is
 $ \dfrac{0.32}{0.26}=\dfrac{16}{13}=\dfrac{48}{39}<\dfrac{52}{39}=\dfrac{4}{3} $
Al is the limiting reagent since its ratio to diatomic oxygen is smaller than $ \dfrac{4}{3} $ .
 $ \text{0}\text{.26 moles }{{\text{O}}_{\text{2}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{4 moles Al}}{\text{3 moles }{{\text{O}}_{\text{2}}}}\approx \text{0}\text{.347 moles Al} $
Hence it would need 0.347 moles of Al to totally react with all of the oxygen, but there are only 0.32 moles of Aluminum present, so there isn't enough Al to do so, and we call Al the limiting reagent.

Note:
First and foremost, the chemical equation must be balanced. The law of conservation says that the amount of each element in a chemical process remains constant. As a result, the chemical equation is balanced when the amounts of each element on both sides of the equation are equal. After that, transform all of the supplied data (usually masses) into moles and compare the mole ratios of the given data to those in the chemical equation.