What is the limiting reactant if \[4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\] \[26.2g\] of \[Fe{S_2}\] react with \[5.44g\] \[{O_2}\] .?

Hint: In a chemical reaction, the limiting reagent is a reactant that is fully absorbed when the reaction is over. This reagent limits the amount of substance produced because the reaction cannot proceed without it. Excess reagents occur when one or more other reagents are present in amounts greater than those needed to react with the limiting reagent.

Complete answer:
The balanced equation is given, that is,
\[4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}\]
To find the limiting reactant, we must calculate the moles of the reactants from the amount we have, which is given in the question, and from the molar mass.
The molar mass of \[Fe{S_2}\] is \[120g\] .
Moles of \[Fe{S_2} = \dfrac{{gramFe{S_2}\left( {given} \right)}}{{\dfrac{g}{{mol}}Fe{S_2}}}\]
That is, moles of \[Fe{S_2} = \dfrac{{26.2g}}{{120\dfrac{g}{{mol}}}} = 0.218mol\]
The molar mass of \[{O_2}\] is \[32g\] .
Moles of \[{O_2} = \dfrac{{gram{O_2}\left( {given} \right)}}{{\dfrac{g}{{mol}}{O_2}}}\]
That is, moles of \[{O_2} = \dfrac{{5.44g}}{{32\dfrac{g}{{mol}}}} = 0.17mol\]
According to the balanced chemical equation, \[4mol\] \[Fe{S_2}\] requires \[11mol\] \[{O_2}\] for the reaction. Hence, for \[0.218mol\] of \[Fe{S_2}\] , the moles of \[{O_2}\] required would be
\[0.218 \times \dfrac{{11}}{4} = 0.60mol\] of \[{O_2}\] .
But we have only \[0.17mol\] \[{O_2}\] . Hence \[{O_2}\] is the limiting reagent in this case.

Additional Information:
Keep in mind that the limiting reagent is the one that is fully consumed. Since the theoretical yield is known as the amount of product obtained when the limiting reagent reacts fully, the limiting reagent must be established in order to determine the percentage yield of a reaction.

Note:
We can find out real-life examples for limiting reagents. That is, if you have \[8\] cars with no wheels and \[48\] tyres and want to put tyres on them, cars would be your restricting reagent because in total \[8\] cars needs \[32\] tyres. You will put tyres on all of the vehicles, and if there are any excess tyres, the cars will be the limiting reagent.