Limiting Reactants and Percent Yield
Identify limiting and excess reactants, calculate theoretical yield, actual yield, and percent yield, and interpret sources of experimental loss.
A lesson of Chemistry, in Chemical Reactions and Stoichiometry, about 35 minutes.
By the end of this lesson, you will be able to find which reactant runs out first, compute the product from that one alone, find how much of the other is left over, and calculate percent yield.
One Ingredient Runs Out First
You have $10$ slices of bread and $3$ slices of cheese, and a sandwich takes two bread and one cheese. You can make three sandwiches, and four slices of bread are left over.
The cheese is limiting. It ran out first and it alone decided the outcome. The bread is in excess, and the amount of bread you had was irrelevant to the answer.
$$2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}$$
Given $10$ moles of hydrogen and $2$ moles of oxygen, the oxygen supports only $4$ moles of water and then stops. Six moles of hydrogen sit unused.
The limiting reactant is not the one you have less of. That is the intuition to unlearn. It is the one that runs out first given the RATIO the equation demands, and $10$ against $2$ still leaves oxygen limiting because the equation wants hydrogen two-to-one.
Everything downstream comes from the limiting reactant, and using the other one produces an answer that is too large and looks fine.
Finding It: Compute Product From Each
The reliable method has no judgment in it at all.
- Convert both reactant masses to moles
- Calculate how much product each one alone would make
- The smaller answer is correct, and that reactant is limiting
The smaller result is the real one because the reaction physically stops when the first reactant is exhausted. The larger figure describes a reaction that could not finish.
$$4.0 \text{ g } \mathrm{H_2} \Rightarrow 2.0 \text{ mol} \Rightarrow 2.0 \text{ mol } \mathrm{H_2O}$$
$$16.0 \text{ g } \mathrm{O_2} \Rightarrow 0.50 \text{ mol} \Rightarrow 1.0 \text{ mol } \mathrm{H_2O}$$
Oxygen gives less, so oxygen is limiting and the answer is $1.0$ mol of water.
A faster alternative is to divide each reactant's moles by its coefficient and take the smaller quotient. It gives the same answer and is easier to misapply, so the two-calculations method is worth using while the idea is new.