14.3: Limiting Reactant
Lesson Objectives
The student will:
 explain what is meant by the terms “limiting reactant” and “excess reactant.”
 determine the limiting reactant when given the mass or the number of moles of the reactants.
 calculate the quantities of products by using the quantity of limiting reactant and molemole ratios.
Vocabulary
 excess reactant
 limiting reactant
Introduction
Suppose you were in the business of building tricycles from the components of frames, handlebars, seats, and wheels. After you completed constructing the tricycle, you boxed it up and shipped it. Suppose further that your supply of components consisted of 100 frames, 120 sets of handlebars, 80 seats, and 300 wheels. What is the maximum number of tricycles you could build and ship? You could build a maximum of 80 tricycles because when you run out of seats, you can no longer complete any more tricycles. The number of tricycles you could build will be limited by the number of seats you have available. In other words, the seats are the limiting component in your tricycle construction. The other components, the ones you have more than enough of, are said to be in excess. If you received a shipment of 100 more seats, could you build 100 more tricycles? No, because now one of the other components would become the limiting component.
In the chemical reactions that you have been working with, when you were given the amount of one substance involved in the reaction, it was assumed that you had exactly the required amount of all the other reagents. In this next section, you will be given the amounts of all the reagents, but not in the correct reacting amounts. You will need to determine which of the reactants limits the amount of product that can be formed, as well as how much product can be produced.
Limiting and Excess Reactants
The limiting reactant is the one that is used up first in the reaction. Consequently, limiting reactants determine the maximum amount of product that can be formed. An excess reactant is any reactant present in an amount that is more than enough to react with the limiting reactant. Some excess reactant remains after the limiting reactant has been used up. Chemists often add one reactant in excess in order to make sure that the limiting reactant is completely consumed. For example, if we wanted to find out how much \begin{align*}\text{CO}_{2(g)}\end{align*} we would obtain if we combusted an aspirin tablet, we would use excess oxygen to make sure that the entire aspirin tablet reacted.


 \begin{align*}\text{C}_9\text{H}_8\text{O}_{4(s)} + 9 \ \text{O}_{2(g)} \rightarrow 9 \ \text{CO}_{2(g)} + 4 \ \text{H}_2\text{O}_{(l)}\end{align*}

The limiting reagent is the aspirin tablet because it is used up in the experiment. Oxygen is the excess reactant.
Let’s look at another example. When you take a piece of aluminum foil and place it into a solution of copper(II) chloride, heat and gas are given off. When the reaction is completed, the blue color of the copper(II) chloride solution fades and a brownish solid (copper) is produced. What is the maximum number of grams of copper that can be produced when \begin{align*}53.96\end{align*} grams of aluminum are reacted with \begin{align*}134.4\end{align*} grams of copper(II) chloride?
The reaction is:


 \begin{align*}2 \ \text{Al}_{(s)} + 3 \ \text{CuCl}_{2(aq)} \rightarrow 3 \ \text{Cu}_{(s)} + 2 \ \text{AlCl}_{3(aq)}\end{align*}

Since \begin{align*}53.96\end{align*} grams of \begin{align*}\text{Al}\end{align*} is equivalent to 2 moles of \begin{align*}\text{Al}\end{align*} and \begin{align*}134.44\end{align*} grams of \begin{align*}\text{CuCl}_2\end{align*} is equivalent to 3 moles, we know that both reactants will be completely used up. As a result, 3 moles of \begin{align*}\text{Cu}\end{align*}, or \begin{align*}187.3\end{align*} grams, will be produced.
If we only had \begin{align*}10.0\end{align*} grams of \begin{align*}\text{Al}\end{align*} instead, what is the maximum number of grams of copper that can be produced? Assuming we still have \begin{align*}134.44\end{align*} grams of \begin{align*}\text{CuCl}_2\end{align*}, then it is obvious that the \begin{align*}\text{Al}\end{align*} is the limiting reactant. By doing a massmass calculation based on the amount of the limiting reagent, we can determine that \begin{align*}35.5\end{align*} grams of \begin{align*}\text{Cu}\end{align*} will be produced.
Consider another example. Assume that we have \begin{align*}72.0 \ \text{g}\end{align*} of water reacting with \begin{align*}190. \ \text{g}\end{align*} of \begin{align*}\text{TiCl}_4\end{align*}.
The balanced equation is:


 \begin{align*}\text{TiCl}_{4(s)} + 2 \ \text{H}_2\text{O}_{(l)} \rightarrow \text{TiO}_{2(s)} + 4 \ \text{HCl}_{(aq)}\end{align*}

We can see from the balanced equation that one mole of \begin{align*}\text{TiCl}_4\end{align*} reacts with two moles of water. The \begin{align*}190. \ \text{g}\end{align*} of \begin{align*}\text{TiCl}_4\end{align*} is exactly one mole. Therefore, \begin{align*}\text{TiCl}_4\end{align*} will react with two moles, or \begin{align*}36.0 \ \text{g}\end{align*} of water. In this reaction, however, we are given \begin{align*}72.0 \ \text{g}\end{align*} of water. Therefore, the water is in excess and \begin{align*}36.0 \ \text{g}\end{align*} of water will be left over when the reaction is complete.
Example:
You are given \begin{align*}34.0 \ \text{g}\end{align*} of \begin{align*}\text{NH}_3\end{align*} and \begin{align*}36.5 \ \text{g}\end{align*} of \begin{align*}\text{HCl}\end{align*}, which react according to the following balanced equation. Determine the limiting reactant and the reactant in excess.


 \begin{align*}\text{NH}_{3(g)} + \ \text{HCl}_{(aq)} \rightarrow \text{NH}_4\text{Cl}_{(s)}\end{align*}

According to the balanced equation, one mole of \begin{align*}\text{NH}_3\end{align*} reacts with one mole of \begin{align*}\text{HCl}\end{align*}. The \begin{align*}34.0 \ \text{g}\end{align*} of \begin{align*}\text{NH}_3\end{align*} is equivalent to two moles, and the \begin{align*}36.5 \ \text{g}\end{align*} of \begin{align*}\text{HCl}\end{align*} is equivalent to one. Since they react with a ratio of \begin{align*}1:1\end{align*}, one mole of \begin{align*}\text{HCl}\end{align*} will react with one mole of \begin{align*}\text{NH}_3\end{align*}. One mole of \begin{align*}\text{NH}_3\end{align*} will be left over. As a result, \begin{align*}\text{HCl}\end{align*} is the limiting reactant, and \begin{align*}\text{NH}_3\end{align*} is in excess.
A straightforward way to find the limiting reactant is as follows:
Step 1: Write out the balanced equation.
Step 2: Convert the given amounts of all reactants into moles.
Step 3: Divide the moles of each reactant by its coefficient in the balanced equation. The component with the lowest resulting number is the limiting reactant.
The first two steps might seem obvious, but why do we need to divide by the coefficients? Consider the tricycle analogy from the beginning of this section. Let's say you had 105 frames, 120 handlebars, 100 seats, and 270 wheels. How many tricycles could you build? Even though you have more wheels than any other component, it is still the limiting factor for this problem, because you need three wheels for each tricycle. Thus, the answer would be 90. If we were to write the “synthesis” of a tricycle as a chemical equation, it would read:


 \begin{align*}\text{frame} + \ \text{handlebars} + \ \text{seat} + 3 \ \text{wheel} \rightarrow \text{tricycle}\end{align*}

If we divide the number of components we have by the coefficients in this “reaction,” we would have the values 105 frames, 120 handlebars, 100 seats, and 90 wheels. Comparing these numbers allow you to quickly see that the wheels are the limiting “reactant.”
Let us look at some examples using real chemical reactions.
Example:
Andrew was working in the lab and mixed a solution containing \begin{align*}25.0 \ \text{g of HCl}\end{align*} with a solution containing \begin{align*}25.0 \ \text{g}\end{align*} of iron(II) sulfide. What is the limiting reagent, and what reagent was in excess?
Step 1: Write the balanced equation:


 \begin{align*}\text{FeS} + 2 \ \text{HCl} \rightarrow \text{H}_2\text{S} + \ \text{FeCl}_2\end{align*}

Step 2: Convert given amounts to moles:


 \begin{align*}\text{mol FeS:} \ \frac {25.0 \ \text{g}} {87.9 \ \text{g/mol}} = 0.284 \ \text{mol FeS}\end{align*}



 \begin{align*}\text{mol HCl:} \ \frac {25.0 \ \text{g}} {36.5 \ \text{g/mol}} = 0.686 \ \text{mol HCl}\end{align*}

Step 3: Divide each reactant by its coefficient in the balanced equation:


 \begin{align*}\text{FeS:} \ \frac {0.284 \ \text{mol}} {1} = 0.284 \ \text{mol}\end{align*}



 \begin{align*}\text{HCl:} \ \frac {0.686 \ \text{mol}} {2} = 0.343 \ \text{mol}\end{align*}

FeS has the lowest final value, so it is the limiting reagent.
Example:
A student mixed \begin{align*}5.0 \ \text{g}\end{align*} of carbon with \begin{align*}23.0 \ \text{g}\end{align*} iron(III) oxide. What is the limiting reactant, and what reactant was the excess reactant?
Step 1: Write the balanced equation:


 \begin{align*}2 \ \text{Fe}_2\text{O}_3 + 3 \ \text{C} \rightarrow 4 \ \text{Fe} + 3 \ \text{CO}_2\end{align*}

Step 2: Convert given amounts to moles:


 \begin{align*}\text{mol C:} \ \frac {5.0 \ \text{g}} {12.0 \ \text{g/mol}} = 0.417 \ \text{mol C}\end{align*}



 \begin{align*}\text{mol Fe}_2\text{O}_3\text{:} \ \frac {23.0 \ \text{g}} {159.7 \ \text{g/mol}} = 0.144 \ \text{mol Fe}_2\text{O}_3\end{align*}

Step 3: Divide each reactant by its coefficient in the balanced equation:


 \begin{align*}\text{C:} \ \frac {0.417 \ \text{mol}} {3} = 0.139 \ \text{mol}\end{align*}



 \begin{align*}\text{mol Fe}_2\text{O}_3\text{:} \ \frac {0.144 \ \text{mol}} {2} = 0.0720 \ \text{mol}\end{align*}

\begin{align*}\text{mol Fe}_2\text{O}_3\end{align*} has the lowest final value, so it is the limiting reagent.
Example:
While working in the lab, Brenda added \begin{align*}5.0 \ \text{g}\end{align*} of aluminum chloride to \begin{align*}7.0 \ \text{g}\end{align*} of silver nitrate in order to work on her double replacement reactions. Her first goal was to determine the limiting reagent. Can you help her determine which reactant is limiting, and which is in excess given the balanced equation?
Step 1: Write the balanced equation:


 \begin{align*}\text{AlCl}_3 + 3 \ \text{AgNO}_3 \rightarrow 3 \ \text{AgCl} + \ \text{Al(NO}_3)_3\end{align*}

Step 2: Convert given amounts to moles:


 \begin{align*}\text{mol AlCl}_3\text{:} \ \frac {5.0 \ \text{g}} {133 \ \text{g/mol}} = 0.037 \ \text{mol AlCl}_3\end{align*}



 \begin{align*}\text{mol AgNO}_3\text{:} \ \frac {7.0 \ \text{g}} {170. \ \text{g/mol}} = 0.041 \ \text{mol AgNO}_3\end{align*}

Step 3: Divide each reactant by its coefficient in the balanced equation:


 \begin{align*}\text{mol AlCl}_3\text{:} \ \frac {0.037 \ \text{mol}} {1} = 0.037 \ \text{mol}\end{align*}



 \begin{align*}\text{mol AgNO}_3\text{:} \ \frac {0.041 \ \text{mol}} {3} = 0.014 \ \text{mol}\end{align*}

\begin{align*}\text{AgNO}_3\end{align*} has the lowest final value, so it is the limiting reagent. Note that this is the case even though there are fewer moles of \begin{align*}\text{AlCl}_3\end{align*}. However, since three units of \begin{align*}\text{AgNO}_3\end{align*} are used up for each unit of \begin{align*}\text{AlCl}_3\end{align*}, it is the \begin{align*}\text{AgNO}_3\end{align*} that will run out first.
Limiting Reactant and MassMass Calculations
Once we have determined the limiting reactant, we still need to determine the amount of product the limiting reactant will produce. We might also want to know how much of the other reactant we are using and how much is left over. Consider again the example where the student mixed 5.0 g carbon and 23.0 g iron(III) oxide. The reaction is:


 \begin{align*}2 \ \text{Fe}_2\text{O}_3 + 3 \ \text{C} \rightarrow 4 \ \text{Fe} + 3 \ \text{CO}_2\end{align*}

We found that \begin{align*}\text{Fe}_2\text{O}_3\end{align*} is the limiting reagent and that \begin{align*}23.0 \ \text{g}\end{align*} of \begin{align*}\text{Fe}_2\text{O}_3\end{align*} is equivalent to \begin{align*}0.144 \ \text{mol}\end{align*}). In order to determine the mass of iron produced, we simply have to perform two additional steps: convert moles of limiting reactant to moles of product (using molar ratios), and convert moles of product to mass (using mass = moles \begin{align*}\times\end{align*} molar mass).
The mass of iron produced is then:


 \begin{align*}\text{mol Fe} = (0.144 \ \text{mol Fe}_2\text{O}_3) \cdot ( \frac {4 \ \text{mol Fe}} {2 \ \text{mol Fe}_2\text{O}_3}) = 0.288 \ \text{mol Fe}\end{align*}



 \begin{align*}\text{mass Fe} = 0.288 \ \text{mol Fe} \cdot 55.85 \ \text{g/mol} = 16.1 \ \text{g Fe}\end{align*}

Therefore, the mass of \begin{align*}\text{Fe}\end{align*} formed from this reaction would be \begin{align*}16.1 \ \text{g}\end{align*}.
Now, let’s consider two more example.
Example:
Magnesium chloride is used as a fireproofing agent, but it is also a coagulant in tofu – imagine that! Jack was working in the lab one day and mixed \begin{align*}1.93 \ \text{g}\end{align*} of magnesium oxide with \begin{align*}3.56 \ \text{g}\end{align*} of HCl. The equation is given below. What mass of magnesium chloride was formed? (Remember to balance the equation first!)


 \begin{align*}\text{MgO} + \ \text{HCl} \rightarrow \text{MgCl}_2 + \ \text{H}_2\text{O}\end{align*}

Step 1: Write the balanced equation.


 \begin{align*}\text{MgO} + 2 \ \text{HCl} \rightarrow \text{MgCl}_2 + \ \text{H}_2\text{O}\end{align*}

Step 2: Convert given amounts to moles.


 \begin{align*}\text{mol MgO:} \ \frac {1.93 \ \text{g}} {40.3 \ \text{g/mol}} = 0.0479 \ \text{mol MgCl}_2\end{align*}



 \begin{align*}\text{mol HCl:} \ \frac {3.56 \ \text{g}} {36.5 \ \text{g/mol}} = 0.0975 \ \text{mol HCl}\end{align*}

Step 3: Divide each amount by the coefficients from the balanced equation.


 \begin{align*}\text{MgO:} \ \frac {0.0479 \ \text{mol}} {1} = 0.0479 \ \text{mol}\end{align*}



 \begin{align*}\text{HCl:} \ \frac {0.0975 \ \text{mol}} {2} = 0.0488 \ \text{mol}\end{align*}

\begin{align*}\text{MgO}\end{align*} has the lowest final value, so it is the limiting reagent. Thus, we use the values for \begin{align*}\text{MgO}\end{align*} in our stoichiometric calculations.
Step 4: Convert moles of the limiting reagent to mass of the desired product.


 \begin{align*}\text{mol MgCl}_2\text{:} \ (0.0479 \ \text{mol MgO}) \cdot (\frac {1 \ \text{mol MgCl}_2} {1 \ \text{mol MgO}} = 0.0479 \ \text{mol MgCl}_2\end{align*}



 \begin{align*}\text{mass MgCl}_2\text{:} \ (0.0479 \ \text{mol MgCl}_2) \cdot (95.2 \ \text{g/mol}) = 4.56 \ \text{g MgCl}_2\end{align*}

Example:
In the following reaction, \begin{align*}12.8 \ \text{g}\end{align*} of ammonia and \begin{align*}14.5 \ \text{g}\end{align*} of oxygen are allowed to react. How many grams of nitrogen monoxide are formed?
Step 1: Write the balanced equation.


 \begin{align*}4 \ \text{NH}_3 + 5 \ \text{O}_2 \rightarrow 4 \ \text{NO} + 6 \ \text{H}_2\text{O}\end{align*}

Step 2: Convert given amounts to moles.


 \begin{align*}\text{mol NH}_3\text{:} \ \frac {12.8 \ \text{g}} {17.0 \ \text{g/mol}} = 0.752 \ \text{mol NH}_3\end{align*}



 \begin{align*}\text{mol O}_2\text{:} \ \frac {14.5 \ \text{g}} {32.0 \ \text{g/mol}} = 0.453 \ \text{mol O}_2\end{align*}

Step 3: Divide each amount by the coefficients from the balanced equation.


 \begin{align*}\text{NH}_3\text{:} \ \frac {0.752 \ \text{mol}} {4} = 0.188 \ \text{mol}\end{align*}



 \begin{align*}\text{O}_2\text{:} \ \frac {0.453 \ \text{mol}} {5} = 0.091 \ \text{mol}\end{align*}

\begin{align*}\text{O}_2\end{align*} has the lowest final value, so it is the limiting reagent. Thus, we use the values for \begin{align*}\text{O}_2\end{align*} in our stoichiometric calculations.
Step 4: Convert moles of the limiting reagent to mass of the desired product.


 \begin{align*}\text{mol NO:} \ (0.453 \ \text{mol O}_2) \cdot (\frac {4 \ \text{mol NO}} {5 \ \text{mol O}_2} = 0.362 \ \text{mol NO}\end{align*}



 \begin{align*}\text{mass NO:} \ (0.362 \ \text{mol NO}) \cdot (30.01 \ \text{g/mol}) = 10.9 \ \text{g NO}\end{align*}

Lesson Summary
 The limiting reactant is the one that is used up first in the reaction.
 The limiting reactant limits the amount of product that can form from a chemical reaction.
 An excess reactant is any reactant present in an amount that is more than enough to react with the limiting reactant.
 The excess reactant remains after the limiting reactant has been used up.
Further Reading / Supplemental Links
This website has several video lessons on performing stoichiometry calculations, including problems with a limiting reactant.
Review Questions
 Consider the balanced reaction: \begin{align*}2 \ \mathrm{Al} + 6 \ \mathrm{HBr} \rightarrow 2 \ \text{AlBr}_3 + 3 \ \text{H}_2\end{align*}.
 When \begin{align*}3.22\ \mathrm{mol}\end{align*} of Al reacts with \begin{align*}4.96\ \mathrm{mol}\end{align*} of \begin{align*}\text{HBr}\end{align*}, how many moles of \begin{align*}\text{H}_2\end{align*} are formed?
 What is the limiting reactant?
 For the reactant in excess, how many moles are left over at the end of the reaction?
 Write the balanced equation for this reaction: copper(II) chloride reacts with sodium nitrate to form copper(II) nitrate and sodium chloride.
 If \begin{align*}15.0 \ \mathrm{g}\end{align*} of copper(II) chloride react with \begin{align*}20.0\ \mathrm{g}\end{align*} of sodium nitrate, how much sodium chloride can be formed?
 What is the limiting reactant for this reaction?
 How much of the nonlimiting reactant will be left over in this reaction?