Solveeit Logo

Question

Question: 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...

What is the limiting reactant if 4FeS2+11O22Fe2O3+8SO24Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2} 26.2g26.2g of FeS2Fe{S_2} react with 5.44g5.44g O2{O_2} .?

Explanation

Solution

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,
4FeS2+11O22Fe2O3+8SO24Fe{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 FeS2Fe{S_2} is 120g120g .
Moles of FeS2=gramFeS2(given)gmolFeS2Fe{S_2} = \dfrac{{gramFe{S_2}\left( {given} \right)}}{{\dfrac{g}{{mol}}Fe{S_2}}}
That is, moles of FeS2=26.2g120gmol=0.218molFe{S_2} = \dfrac{{26.2g}}{{120\dfrac{g}{{mol}}}} = 0.218mol
The molar mass of O2{O_2} is 32g32g .
Moles of O2=gramO2(given)gmolO2{O_2} = \dfrac{{gram{O_2}\left( {given} \right)}}{{\dfrac{g}{{mol}}{O_2}}}
That is, moles of O2=5.44g32gmol=0.17mol{O_2} = \dfrac{{5.44g}}{{32\dfrac{g}{{mol}}}} = 0.17mol
According to the balanced chemical equation, 4mol4mol FeS2Fe{S_2} requires 11mol11mol O2{O_2} for the reaction. Hence, for 0.218mol0.218mol of FeS2Fe{S_2} , the moles of O2{O_2} required would be
0.218×114=0.60mol0.218 \times \dfrac{{11}}{4} = 0.60mol of O2{O_2} .
But we have only 0.17mol0.17mol O2{O_2} . Hence O2{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 88 cars with no wheels and 4848 tyres and want to put tyres on them, cars would be your restricting reagent because in total 88 cars needs 3232 tyres. You will put tyres on all of the vehicles, and if there are any excess tyres, the cars will be the limiting reagent.