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.
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??
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