Limiting Reactant

You will often hear a chemistry student/prof say something like "it's a limiting reactant problem". This means that your reactants are not given in perfect stoichiometric ratios. This means that one of the reactants is "limiting" in that it will limit the amount of product you can make. All the other reactants that aren't limiting are said to be "in excess". You will need to calculate all product quantities based on the limiting reactant and not the ones in excess. So you need to have a method that allows you to figure out which reactant is limiting.

There are multiple ways to go about this. It is easy when your "reaction" is really just making sandwiches for a soccer team. Let's do this exercise for a simple bologna sandwich. One bologna sandwich is two slices of bread and one slice of bologna. There are 11 on the team. You need 11 sandwiches. You do inventory... you have 14 slices of bologna and 18 slices of bread. How many sandwiches can you make? Nine. The bread is limiting you from making more than nine sandwiches. You'll make nine, and there will be no bread slices left, but there will be five bologna slices left over. If you understand that, you have the basis for the limiting reactant concept.

How did you THINK about that sandwich problem? Here is one way... I've got 14 slices of bologna. This means I need 28 slices of bread to match up with those 14 slices. I don't have 28 slices, I only have 18 slices. I conclude - the bread slices will limit me and therefore I can only make 9 sandwiches.

Another way I notice I have 18 slices of bread. This means that I need 9 slices of bologna to make sandwiches out of all the bread. I have 14 slices of bologna - that is more than enough, as a matter of fact, it is exactly 5 more which means I can make 9 sandwiches and have 5 leftover slices of bologna.

Now lets work a chemistry problem that is a limiting reactant problem. By-the-way, most limiting reactant problems aren't flagged as such - they're just a stoichiometry problem, it is up to YOU to recognize that it is a limiting reactant problem and treat it as such.

Limiting Reactant Question

Consider the following reaction of methane with chlorine gas to make carbon tetrachloride and hydrogen chloride.

CH₄ + 4Cl₂ → CCl₄ + 4HCl

(a) How many grams of HCl will be produced when 97.2 g of chlorine gas is allowed to completely react with 7.25 g of methane gas? (b) Which reactant (if any) is a leftover after the reaction and how much is there?

Solution (a): Let's first convert the methane into chlorine and see if we have enough or too much:

\(\newcommand\ccancel[2][black]{\color{#1}{\bcancel{\color{black}{#2}}}} \left({7.25\,\ccancel[red]{\rm g\;CH_4}\over 1}\right) \left({1\,\ccancel[green]{\rm mol\;CH_4}\over 16.0\,\ccancel[red]{\rm g\;CH_4}}\right) \left({4\,\ccancel[blue]{\rm mol\;Cl_2}\over 1\,\ccancel[green]{\rm mol\;CH_4}}\right)\) \(\left({70.9\,{\rm g\;Cl_2}\over 1\,\ccancel[blue]{\rm mol\;Cl_2}}\right) = 128.5\,{\rm g\;Cl_2\;needed}\)

We were given 97.2 g of Cl₂ which is less than the needed 128.5 g. This means that the Cl₂ is the limiting reactant and we should use it to calculate the amount of HCl produced.

\(\left({97.2\,\ccancel[red]{\rm g\;Cl_2}\over 1}\right) \left({1\,\ccancel[green]{\rm mol\;Cl_2}\over 70.9\,\ccancel[red]{\rm g\;Cl_2}}\right) \left({4\,\ccancel[blue]{\rm mol\;HCl}\over 4\,\ccancel[green]{\rm mol\;Cl_2}}\right)\) \(\left({36.5\,{\rm g\;HCl}\over 1\,\ccancel[blue]{\rm mol\;HCl}}\right) = 50.0\,{\rm g\;HCl\;produced}\)

So we have our answer to the first part of the question - we produce 50 g of HCl.

Solution (b) Now to get the second part of the question we need to calculate how much of the methane does react and how much is left over. We basically do the same thing as we did previous but get the stoichiometric amount of methane that reacts from the amount of chlorine given

\(\left({97.2\,\ccancel[red]{\rm g\;Cl_2}\over 1}\right) \left({1\,\ccancel[green]{\rm mol\;Cl_2}\over 70.9\,\ccancel[red]{\rm g\;Cl_2}}\right) \left({1\,\ccancel[blue]{\rm mol\;CH_4}\over 4\,\ccancel[green]{\rm mol\;Cl_2}}\right)\) \(\left({16.0\,{\rm g\;CH_4}\over 1\,\ccancel[blue]{\rm mol\;CH_4}}\right) = 5.48\,{\rm g\;CH_4\;reacts}\)

Now to get the leftovers we just subtract from the original amount (excess) given:

\(7.25\;{\rm g} - 5.48\;{\rm g} = 1.77\;{\rm g\;leftover\;CH_4}\)

Percent Yield

Many chemical reactions do not completely go 100% forward according to theory and our lovely balanced equation. Sometimes there are complications and other side reactions may occur. Or, maybe the reaction just doesn't completely push forward as our balanced reaction indicates - maybe a portion of the reactants is left behind unreacted. We can easily show just how far forward a reaction goes by reporting the percent yield of the reaction. This is just a percentage that shows the actual yield (what you actually got and measured) vs the theoretical yield (the amount you get if ALL the reactants react as far as they can) shown as a percentage. Realize that the theoretical yield is 100% yield and is the highest you can get. Here is the formula:

percent yield = actual yield theoretical yield × 100%

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