If you have the equation of a reaction, and are told that there is 100grams of each of the reactant, what do you do in order to find which reactant will limit the reaction?

Limiting reagent problems are all about moles, more specifically about moles of reactants.

Two two things to look out for in such cases are

• how many moles of each reactant would you need in order for the reaction to take place
• how many moles of each reactant you actually have

Your tool of choice for limiting reagent problems, and for any stoichiometry problme, for that matter, is the mole ratio, that is, the ratio between how many moles of each reatant are needed in order for the reaction to take place.

Let's say that you have a generic reaction

#"A" + color(red)(2)"B" -> "C" + "D"#

The stoichiometric coefficients that are placed in front of each reactant represent how many moles of each are needed. In this case, you have #1# mole of reactant #"A"# and #color(red)(2)# moles of reactant #"B"#.

This means that you have a #1: color(red)(2)# mole ratio between reactant #"A"# and reactant #"B"#.

Simply put, regardless of how many moles of #"A"# you have, you will always need twice as many moles of #"B"#.

With this in mind, you would go on to calculate how many moles of each reactant you have by suing their respective molar mass.

#"no. of moles" = "mass"/"molar mass"#

Once you have the number of moles of each reactant, compare them with the #1:color(red)(2)# mole ratio.

Here's how you would do that. Let's say that those #"100-g"# samples contain #1.5# moles of #"A"# and #2.5# moles of #"B"#, respectively.

Pick one of the two rectants, and check to see if you have enough number of moles of the other reactant. Let's pick #"A"#

#1.5color(red)(cancel(color(black)("moles of A"))) * (color(red)(2)" moles of B")/(1color(red)(cancel(color(black)("mole of A")))) = "3 moles of B"#

This tells you that in order for all the moles of #"A"# to react, you would need #3# moles of #"B"#. Since you only have #2.5# moles of #"B"#, it follows that #"B"# will be the limiting reagent.

Alternatively, we could have picked #"B"#

#2.5color(red)(cancel(color(black)("moles of B"))) * ("1 mole of A")/(color(red)(2)color(red)(cancel(color(black)("moles of B")))) = "1.25 moles of A"#

In order for all the number of moles of #"B"# to rect, you would need #1.25# moles of #"A"#. Since you have #1.5# moles of #"A"#, it follows that #"A"# is in excess, which is another way of saying that #"B"# is the limiting reagent.

So, as a conclusion, in order to determine the limiting reagent, you must

• determine the mole ratio that exists between the reactants by looking at the balanced chemical equation
• use the rectants' respective molar masses to find how many moles of each you have
• compare the resulting number of moles with the mole ratio by picking a reactant and checking to see if you have enough moles of the second reactant