Limiting Reagents

Optional Activity: What is a limiting reagent?

For this activity you will need A4 sheets of paper in white, red, blue, yellow, green and pink. (or you can use several sheets of white paper and colour them using kokis or crayons).

Tear the white sheet into five pieces, the red sheet into ten pieces, the blue sheet into eight pieces, the yellow sheet into seven pieces, the green sheet into nine pieces and the pink piece into four pieces.

Stick two red pieces to each white piece. Do you have any red or white pieces left?
Stick one yellow piece to each blue piece. Do you have any yellow or blue pieces left?
Stick three green pieces to each pink piece. Do you have any green or pink pieces left?

You should find that you had no red or white pieces left. For the blue and yellow pieces you should have one blue piece left. And for the green and pink pieces you should have had one pink piece left.

We say that the pink and blue pieces were in excess while the green and yellow sheets were limiting. In other words you would have had to tear the green and yellow sheets into more pieces or you would have had to tear the blue and pink pieces into less pieces.

In the above activity we could solve the problem of having too many or too few pieces of paper by simply tearing the pieces of paper into more pieces. In chemistry we also encounter this problem when mixing different substances. Often we will find that we added too much or too little of a particular substance. It is important to know that this happens and to know how much (i.e. the quantities) of different reactants are used in the reaction. This knowledge is used in industrial reactions.

Definition: Limiting reagent

A limiting reagent (or reactant) is a reagent that is completely used up in a chemical reaction.

Definition: Excess reagent

An excess reagent (or reactant) is a reagent that is not completely used up in a chemical reaction.

PhET Simulation: Reactants, Products and Leftovers

Example: Limiting Reagents

Question

Sulfuric acid (\(\text{H}_{2}\text{SO}_{4}\)) reacts with ammonia (\(\text{NH}_{3}\)) to produce the fertiliser ammonium sulfate (\((\text{NH}_{4})_{2}\text{SO}_{4}\)) according to the following equation:

\(\text{H}_{2}\text{SO}_{4}\text{(aq)} + 2\text{NH}_{3}\text{(g)} → \text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\text{(aq)}\)

What is the maximum mass of ammonium sulfate that can be obtained from \(\text{2.0}\) \(\text{kg}\) of sulfuric acid and \(\text{1.0}\) \(\text{kg}\) of ammonia?

Step 1: Convert the mass of sulfuric acid and ammonia into moles

Moles of sulfuric acid: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{2 000}}{\text{98}} \\ & = \text{20.4}\text{ mol} \end{align*}

Moles of ammonia: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{1 000}}{\text{17}} \\ & = \text{58.8}\text{ mol} \end{align*}

Step 2: Use the balanced equation to determine which of the reactants is limiting.

We need to look at how many moles of product we can get from each reactant. Then we compare these two results. The smaller number is the amount of product that we can produce and the reactant that gives the smaller number, is the limiting reagent.

The mole ratio of \(\text{H}_{2}\text{SO}_{4}\) to \((\text{NH}_{4})_{2}\text{SO}_{4}\) is \(1:1\). So the number of moles of \((\text{NH}_{4})_{2}\text{SO}_{4}\) that can be produced from the sulfuric acid is: \begin{align*} n_{\text{(NH}_{4}\text{)}_{2}\text{SO}_{4}} & = n_{\text{H}_{2}\text{SO}_{4}} \times \frac{\text{stoichiometric coefficient (NH}_{4}\text{)}_{2}\text{SO}_{4}}{\text{stoichiometric coefficient H}_{2}\text{SO}_{4}}\\ & = \text{20.4}\text{ mol} \text{ H}_{2}\text{SO}_{4} \times \frac{\text{1}\text{ mol} \text{ (NH}_{4}\text{)}_{2}\text{SO}_{4}}{\text{1}\text{ mol} \text{ H}_{2}\text{SO}_{4}}\\ & = \text{20.4}\text{ mol} \text{ (NH}_{4}\text{)}_{2}\text{SO}_{4} \end{align*}

The mole ratio of \(\text{NH}_{3}\) to \((\text{NH}_{4})_{2}\text{SO}_{4}\) is \(2:1\). So the number of moles of \((\text{NH}_{4})_{2}\text{SO}_{4}\) that can be produced from the ammonia is: \begin{align*} n_{\text{(NH}_{4}\text{)}_{2}\text{SO}_{4}} & = n_{\text{NH}_{3}} \times \frac{\text{stoichiometric coefficient (NH}_{4}\text{)}_{2}\text{SO}_{4}}{\text{stoichiometric coefficient NH}_{3}}\\ & = \text{58.8}\text{ mol} \text{ NH}_{3} \times \frac{\text{1}\text{ mol} \text{ (NH}_{4}\text{)}_{2}\text{SO}_{4}}{\text{2}\text{ mol} \text{ NH}_{3}}\\ & = \text{29.4}\text{ mol} \text{ (NH}_{4}\text{)}_{2}\text{SO}_{4} \end{align*}

Since we get less \((\text{NH}_{4})_{2}\text{SO}_{4}\) from \(\text{H}_{2}\text{SO}_{4}\) than is produced from \(\text{NH}_{3}\), the sulfuric acid is the limiting reactant.

Step 3: Calculate the maximum mass of ammonium sulfate that can be produced

From the step above we saw that we have \(\text{20.4}\) \(\text{mol}\) of \((\text{NH}_{4})_{2}\text{SO}_{4}\).

The maximum mass of ammonium sulfate that can be produced is calculated as follows:

\begin{align*} m & = nM\\ & = (\text{20.4})(\text{132}) \\ & = \text{2 692.8}\text{ g}\\ & = \text{2.6928}\text{ kg} \end{align*}

The maximum mass of ammonium sulfate that can be produced is \(\text{2.69}\) \(\text{kg}\).

Limiting Reagents