5.4 Limiting Reagents and Yield

Learning Objectives

  • Utilising stoichiometry coefficients and molar mass calculations, determining limiting reagents.
  • Applying limiting reagents to calculate theoretical yields – allowing for determination of overall yield.

So far, we are yet to consider some of the practical limitations encountered when considering real chemical reactions in the laboratory. We have assumed that reactions proceed all the way to completion and that all reactants are in excess. In the real world, this is not always the case. In this chapter, let’s address those assumptions.

Limiting and Excess Reagents

Let’s consider the reaction between hydrogen gas and nitrogen gas to form ammonia gas: \[

Based on stoichiometry, we know that if we mix three moles of  [latex]\ce{H}_{2}[/latex] with one mole of [latex]\ce{N}_{2}[/latex] we can obtain two moles of  [latex]\ce{NH}_{3}[/latex]. In other words, [latex]\ce{H}_{2}[/latex] reacts with [latex]\ce{N}_{2}[/latex] in a molar ratio of  [latex]3:1[/latex]. If we mix three moles of hydrogen gas and two moles of nitrogen gas, what will happen? Similarly, three moles of  [latex]\ce{H}_{2}[/latex] will react with one mole of [latex]\ce{N}_{2}[/latex] and produce two moles of  [latex]\ce{NH}_{3}[/latex]. However, the entire quantity of [latex]\ce{H}_{2}[/latex] will be used before the [latex]\ce{N}_{2}[/latex] as we start with the exact amount of [latex]\ce{H}_{2}[/latex] required for the reaction. The product mixture will contain one unreacted mole of [latex]\ce{N}_{2}[/latex] as we only required one mole of [latex]\ce{N}_{2}[/latex] for the reaction. If we had more [latex]\ce{H}_{2}[/latex] we could have produced more [latex]\ce{NH}_{3}[/latex] as we already have excess [latex]\ce{N}_{2}[/latex]. Therefore, in this reaction, we call [latex]\ce{H_{2}}[/latex] the limiting reagent since it limits the production of [latex]\ce{NH}_{3}[/latex].

The limiting reagent is the reactant that runs out first, limiting the formation of products. In the above example, nitrogen gas is called the excess reagent because we have a greater amount than required for the reaction.

Example: phosphoric acid is produced by treating phosphate rock, which contains calcium phosphate, with sulphuric acid as follows: \[
A chemist in a fertiliser production plant mixed [latex]144.5\textrm{g}[/latex] of calcium phosphate with [latex]160.5\textrm{g}[/latex] of sulphuric acid. What is the maximum mass of phosphoric acid ([latex]\ce{H}_{3}\ce{PO}_{4}[/latex]) that can be formed? Molar masses are given in table 5.4.1.

Table 5.4.1: Molar masses for production of phosphoric acid
Molecule Molar mass (g/mol)
[latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex] [latex]310.2[/latex]
[latex]\ce{H}_{2}\ce{SO}_{4}[/latex] [latex]98.1[/latex]
[latex]\ce{H}_{3}\ce{PO}_{4}[/latex] [latex]97.99[/latex]

We are given the mass of reactants and are asked to determine the mass of one of the products.

Step 1 – Identify the limiting reagent in this reaction. For that, find the number of moles of each reactant using [latex]n=\frac{m}{M}[/latex].

Number of moles of  [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex]: \[\begin{align*}
n & =\frac{m}{M}\\
n & =\frac{144.5\textrm{g}}{310.2\textrm{ g}/\textrm{mol}}\\
n & =0.4658\textrm{mol}

Number of moles of [latex]\ce{H}_{2}\ce{SO}_{4}[/latex]: \[\begin{align*}
n & =\frac{m}{M}\\
n & =\frac{160.5\textrm{g}}{98.1\textrm{g}/\textrm{mol}}\\
n & =1.64\textrm{mol}

Then, compare the calculated moles with the molar ratio of the two reactants. According to the balanced chemical equation, calcium phosphate reacts with sulphuric acid in a [latex]1:3[/latex] molar ratio. Therefore, [latex]0.4658\textrm{mol}[/latex] of [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex] reacts with the following number of moles of [latex]\ce{H}_{2}\ce{SO}_{4}[/latex]: \[\begin{align*}
\frac{3\textrm{mol}\ce{H}_{2}\ce{SO}_{4}}{1\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}}\times0.4658\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2} & =1.397\textrm{mol}

We have [latex]1.64\textrm{mol}[/latex] of [latex]\ce{H}_{2}\ce{SO}_{4}[/latex], but we only need [latex]1.397\textrm{mol}[/latex] for the reaction. This means we have excess [latex]\ce{H}_{2}\ce{SO}_{4}[/latex] acid. Therefore, the limiting reagent is [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex]. The quantity of limiting reagent can determine the mass of [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid formed.

Step 2 – Use the molar ratio between [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex] and [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid to calculate the mass of [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid formed.

We know one mole of [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex] forms two moles of [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid from the balanced chemical equation. Therefore, the number of moles of [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid forms from [latex]0.4658\textrm{mol}[/latex] of  [latex]\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}[/latex] is: \[\begin{align*}
\frac{2\textrm{mol}\ce{H}_{3}\ce{PO}_{4}}{1\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2}}\times0.4658\textrm{mol}\ce{Ca}_{3}\left(\ce{PO}_{4}\right)_{2} & =0.9316\textrm{mol}

Now, find the mass of  [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] using [latex]M=\frac{m}{n}[/latex] \[\begin{align*}
n & =\frac{m}{M}\\
m & =M\times n\\
m & =97.99\textrm{g}/\textrm{mol}\times0.9316\textrm{mol}\\
m & =91.29\textrm{g}

Based on the given reaction conditions, the maximum mass we can obtain for [latex]\ce{H}_{3}\ce{PO}_{4}[/latex] acid is [latex]91.29\textrm{g}[/latex].

One of the fundamental principles in chemistry and physics is the law of conservation of energy and mass. This famous rule explains that we cannot create nor destroy matter, only transform it. Just like baking a cake, you can only get as much as you put it! The amont of reactants present is our core limitation of how much a reaction can produce – which will form a foundational concept within the topic of yields.


As we have just seen, stoichiometry allows us to determine the maximum mass or amount of a product a reaction can generate for a given quantity of reactants. In reality, chemists never extract the full 100% of a generated product in a reaction.

Cup of coffee on wooden table.
Figure 5.4.1: Notice how the brown staining on the lip of the mug presents. Within it contains caffeine yet to be recovered! Image attribution: A cup of coffee with heart shaped latte design presented on a wooden surface in the morning. © asiraj – stock.adobe.com.

Consider a mug of coffee. Once you have ingested the liquid, there is often a visible brown stain in the mug due to the adhesive forces of the liquid (see Figure 5.4.1). There is likely to be trace amounts of caffeine that you have yet to digest. While you could fill the cup with water and attempt to recover the caffeine, eliminating the visible colour, on the molecular scale with sextillions of molecules, it is impossible to capture every compound. Even in perfect reactions that go to completion with many repetitive washes to gather all of a produced product, we are unable to confidently have recovered everything.

From a more chemical perspective, however, side reactions are a common occurrence in many processes. For example, in the combustion of some carbon-containing compounds, such as methane, there are 2 potential reactions that can occur depending on the available levels of oxygen. They are:

\[\ce{CH}_{4}\left(g\right)+\ce2{O}_{2}\left(g\right)\rightarrow\ce{CO}_{2}\left(g\right)\ce2{H}_{2}{O}\left(g\right)\] \[\ce2{CH}_{4}\left(g\right)+\ce3{O}_{2}\left(g\right)\rightarrow\ce2{CO}\left(g\right)\ce4{H}_{2}{O}\left(g\right)\]

For now, don’t worry about side reactions and equilibriums. We will bring them up if important.

Complete combustion (with oxygen excess) generates [latex]\ce{CO}_{2}[/latex] (carbon dioxide), whereas incomplete combustion produces [latex]\ce{CO}[/latex] (carbon monoxide) – a more toxic greenhouse gas responsible for smog! Car manufacturers attempt to design engines with sufficient airflow to allow for excess oxygen in the environment. However, as they might, some CO is always produced. This, alongside every other chemical reaction, is inefficient to some degree. It is impossible to recover 100% of a reaction.

Finally, we can address our first assumption, that all reactions go to completion. This is not true. While most reactions will undergo enough conversion that we can realistically say they have completed fully, some reactions instead exist in equilibrium, where only a certain amount of potential product is produced. The equation we looked at the beginning of this chapter, the production of ammonia, is one of the most famous equilibrium reactions, known as the Haber process:

[latex]\ce{3H2(g) + N2(g) <=> 2NH3(g)}[/latex]

In the Haber process, reactants and products exist in equilibrium with each other (denoted by the two-way arrow in the equation). This equilibrium can be influenced by air pressure and temperature – which, if carefully managed, allows for the extraction of ammonia. However, this system will never allow for the full 100% extraction of the gas.

This is all to say that in chemical processes, we must consider our percentage yield. Percentage yield indicates the success of the reaction. If the value is closer to 100%, that means losses are minor, and most reactants have converted to products. This is calculated by considering the theoretical yield. as previously demonstrated, and the actual yield, measured in a laboratory, in the following formula:

\frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100

The theoretical yield is the maximum amount of the product that can form under the given reaction conditions. The available amount of limiting reagent determines the maximum amount of the product that can be formed. As demonstrated previously, theoretical yield can be calculated using the stoichiometric coefficients (or molar ratios) displayed in the balanced chemical equation. We calculate theoretical yield assuming that no losses occur during the reaction and all the reactants convert to products. However, in reality, losses and inefficiencies often occur during chemical reactions. Therefore, the actual amount that you isolate and measure from a chemical reaction is less than the theoretically expected amount.

The actual yield of a product is the amount of the product you isolate from the chemical reaction. You cannot calculate the actual yield. It is required to be determined experimentally, generally through weighing the product produced.

From these two values, the percentage yield can be calculated. As a general rule of thumb, yield decreases with the number of steps needed to complete a chemical reaction but can be increased through repeat filtrations and extractions within an experiment. In chemistry and especially industrial applications, bolstering percentage yield is essential to cost-effective operations and the adoption of processes.

Example: When [latex]168.3\textrm{g}[/latex] of [latex]\ce{Fe}[/latex] react with [latex]\ce{O}_{2}[/latex] according to the following chemical equation, [latex]172.6\textrm{g}[/latex] of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] is obtained. What is the percentage yield of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] for this reaction? The molar mass of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] is [latex]159.69\textrm{g}/\textrm{mol}[/latex] and [latex]\ce{Fe}[/latex] is [latex]55.85\textrm{g}/\textrm{mol}[/latex]. \[

We are given the mass of the reactant [latex]\ce{Fe}[/latex], the actual yield of the product [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex], the balanced chemical equation and are asked to calculate the percentage yield of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] for this reaction. We need to find out the theoretical yield of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] for this reaction to calculate the percentage yield.

Step 1 – Use the molar ratios given in the balanced chemical equation and the given mass of [latex]\ce{Fe}[/latex] to calculate the theoretical yield of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex]:.

Number of moles of [latex]\ce{Fe}[/latex]: \[\begin{align*}
n & =\frac{m}{M}\\
n & =\frac{168.3\textrm{g}}{55.85\textrm{g}/\textrm{mol}}\\
n & =3.013\textrm{mol}

From the balanced chemical equation, we know four moles of [latex]\ce{Fe}[/latex] form two moles of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex]. Therefore, [latex]3.013\textrm{mol}[/latex] of [latex]\ce{Fe}[/latex] forms the following amount of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex]: \[\begin{align*}
\frac{2\textrm{mol}}{4\textrm{mol}}\times3.013\textrm{mol} & =1.507\textrm{mol}

Convert the number of moles to [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] mass: \[\begin{align*}
n & =\frac{m}{M}\\
m & =M\times n\\
m & =159.69\textrm{g}/\textrm{mol}\times1.507\textrm{mol}\\
m & =240.7\textrm{g}

Step 2 – Calculate the percentage yield of [latex]\ce{Fe}_{2}\ce{O}_{3}[/latex] using the calculated theoretical yield and the given actual yield: \[\begin{align*}
\frac{\textrm{Actual yield}}{\textrm{Theoretical yield}}\times100 & =\frac{172.6\textrm{g}}{240.7\textrm{g}}\times100=71.71\%

Key Takeaways

  • Reactions are limited by the reactant that runs out first – known as the limiting reagent. This is calculated by comparing the molar ratios against the moles of reactant actually used in the experiment for each substance in a reaction.
  • Theoretical yield is the maximum amount of product that can be generated in a chemical reaction according to the limiting reagent.
  • Experimental yield is the amount of product actually recovered.
  • Percentage yield is the fraction of the experimental yield compared to the theoretical yield.



Practice Questions





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