Suppose 77.4 g P4S6 (MM = 316.24 g/mol) and
269 g KClO2 (MM = 106.55 g/mol) were reacted
together. What is the limiting reagent, and how many
moles of the limiting reagent were initially present?
P4S6 + 11 KClO2 -> 2 P2O5 + 6 SO2 + 11 KCl
We got the answer: 0.25 moles of P4S6 and 2.5 moles of KClO2 but WHY is 2.5 moles of KClO2 the limiting reagent if there are more moles of it??
How much (pick any product--I'll pick SO3) mols SO2 will be formed from 0.25 mols P4S6 if we had all of the KClO2 we needed to react with all of the P4S6? That's 0.25 mols P4S6 x (6 mols SO2/1 mol P4S6) = 0.25 x 6 = 1.5 mols SO2 formed.
Now how many mols SO2 will be formed if we use 2.5 mol KClO2 and all of the P4S6 we need to react with all of the KClO2. That's 2.5 mols KClO2 x (6 mols SO2/11 mols KClO2) = 2.5 x 6/11 = 1.36 mols SO2. So the limiting reagent is the one that produces the FEWEST moles of the product and that's KClO2.
The answer to your question is that the limiting reagent is the one that will produce the fewest moles of any of the products because after the LR is used up, there is no more there to react with the other reagent even though there is some of that reactant left unreacted. I've been doing this homework help for several years now and it seems a favorite for profs to use problems in which the LR is the one with more mols. That's done by using an equation where the coefficients make that happen as in this case. They are trying to trick you into using the smaller moles of reactant because at first glance that appears to be the one to choose. Hope this helps.
To determine the limiting reagent in a chemical reaction, you need to compare the number of moles of each reactant with respect to the stoichiometry of the balanced equation. The stoichiometry tells you the molar ratio of the reactants involved in the reaction.
In the balanced equation:
P4S6 + 11 KClO2 -> 2 P2O5 + 6 SO2 + 11 KCl
The stoichiometric ratio is 1 mole of P4S6 to 11 moles of KClO2. This means that for every 1 mole of P4S6, you need 11 moles of KClO2 for the reaction to proceed optimally.
In this case, you have 77.4 g of P4S6, which can be converted to moles by dividing the given mass by its molar mass:
moles of P4S6 = 77.4 g / 316.24 g/mol = 0.2445 moles
Similarly, you have 269 g of KClO2, which can be converted to moles:
moles of KClO2 = 269 g / 106.55 g/mol = 2.5232 moles
Comparing the number of moles to the stoichiometry of the balanced equation, you can see that there is an excess of KClO2 compared to P4S6. However, the stoichiometric ratio indicates that you need 11 moles of KClO2 for every mole of P4S6.
Since you have more moles of KClO2 than the stoichiometry requires, it is not the limiting reagent. The limiting reagent is the reactant that is completely consumed first, limiting the amount of product that can be formed.
In this case, the limiting reagent is P4S6 because it has fewer moles than KClO2. The 0.2445 moles of P4S6 will react with only 0.2445 moles * 11 moles KClO2/mole P4S6 = 2.6895 moles of KClO2.
Therefore, P4S6 is the limiting reagent, and initially, there were 0.2445 moles of P4S6 present.