Limiting Reactant, Theoretical Yield, and Percent Yield: Step-by-Step Guide

In the laboratory, chemical reactions rarely go exactly according to plan. Just as a baker might run out of eggs before using up all the flour, a chemist often runs out of one specific chemical before the others are consumed. This is the fundamental concept of stoichiometry: using mathematics to predict the outcomes of chemical reactions.

Whether you are calculating the mass of precipitate formed or determining the efficiency of a synthesis, mastery of three core concepts is essential: the limiting reactant, theoretical yield, and percent yield.

This guide breaks down these concepts using real-world reaction data, moving from identifying the bottleneck in your reaction to calculating how efficient your experiment actually was.

Identifying the Limiting Reactant (Reagent)

The limiting reactant (or limiting reagent) is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot proceed without it. The reactant that remains after the reaction stops is called the excess reactant.

Key Definition: The limiting reactant determines the maximum amount of product that can be formed. Once it is gone, the reaction stops, regardless of how much of the other reactants remain.

How to Find the Limiting Reagent

To find the limiting reactant, you cannot simply compare the masses (grams) of the starting materials. You must compare them on a mole-to-mole basis using the balanced chemical equation. Here is the systematic method:

Balance the Chemical Equation: Ensure the stoichiometry is correct.
Convert Masses to Moles: Use molar masses to convert grams of each reactant to moles.
Compare Mole Ratios: Divide the calculated moles of each reactant by its coefficient in the balanced equation. The reactant with the lowest ratio is the limiting reactant.

Case Study: Magnesium Hydroxide and HCl

Let’s look at a neutralization reaction. Suppose we react 50.6 g of Mg(OH)₂ with 45.0 g of HCl. We need to determine which reagent limits the production of Magnesium Chloride ($MgCl_2$).

1. The Balanced Equation:

$$ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O $$

2. Convert to Moles:

Molar Mass of $Mg(OH)_2 \approx 58.33 \text{ g/mol}$
Molar Mass of $HCl \approx 36.46 \text{ g/mol}$

$$ \text{Moles } Mg(OH)_2 = \frac{50.6 \text{ g}}{58.33 \text{ g/mol}} = 0.868 \text{ mol} $$ $$ \text{Moles } HCl = \frac{45.0 \text{ g}}{36.46 \text{ g/mol}} = 1.234 \text{ mol} $$

3. Compare Ratios (Stoichiometric Check):

For $Mg(OH)_2$: $0.868 \text{ mol} / 1 = 0.868$
For $HCl$: $1.234 \text{ mol} / 2 = 0.617$

Since $0.617 < 0.868$, HCl is the limiting reactant. Even though we have a similar mass of both, the reaction requires two moles of HCl for every one mole of Magnesium Hydroxide, causing the acid to run out first.

Calculating Theoretical Yield in Grams

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. It assumes a perfect reaction with no loss. Crucially, you must use the limiting reactant to calculate this value.

Step-by-Step Calculation: Combustion of Butane

Let's calculate the theoretical yield of $CO_2$ when 34.7 g of butane ($C_4H_{10}$) burns in the presence of 55.1 g of oxygen ($O_2$).

Step 1: Balance the equation
Combustion reactions can be tricky to balance. The balanced equation for butane is:

$$ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O $$

Step 2: Determine the Limiting Reactant

Moles $C_4H_{10}$: $34.7 \text{ g} / 58.12 \text{ g/mol} \approx 0.597 \text{ mol}$
Moles $O_2$: $55.1 \text{ g} / 32.00 \text{ g/mol} \approx 1.722 \text{ mol}$

Using the ratio method:

Butane ratio: $0.597 / 2 = 0.298$
Oxygen ratio: $1.722 / 13 = 0.132$

Oxygen has the lower ratio, so $O_2$ is the limiting reactant.

Step 3: Calculate Moles of Product
We must use the stoichiometry between the limiting reactant ($O_2$) and the product ($CO_2$). The ratio is 13:8.

$$ 1.722 \text{ mol } O_2 \times \frac{8 \text{ mol } CO_2}{13 \text{ mol } O_2} = 1.060 \text{ mol } CO_2 $$

Step 4: Convert to Mass (Theoretical Yield)
Finally, convert the moles of $CO_2$ to grams ($MM \approx 44.01 \text{ g/mol}$).

$$ 1.060 \text{ mol} \times 44.01 \text{ g/mol} = 46.6 \text{ g } CO_2 $$

Therefore, the theoretical yield is 46.6 g of $CO_2$.

Determining Percent Yield and Actual Yield

In a real lab setting, you rarely obtain the theoretical yield. Spills, incomplete reactions, or competing side reactions result in a lower amount, known as the actual yield.

The percent yield is a measure of the reaction's efficiency:

$$ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% $$

Example 1: Calculating Percent Yield

Consider the reaction of Copper with Chlorine gas:

$$ Cu + Cl_2 \rightarrow CuCl_2 $$

If 12.5 g of Copper reacts with excess Chlorine, we calculate a theoretical yield of 26.4 g of Copper(II) Chloride ($CuCl_2$). However, after the experiment, only 25.4 g were isolated (Actual Yield).

To find the efficiency:

$$ \text{Percent Yield} = \frac{25.4 \text{ g}}{26.4 \text{ g}} \times 100\% = 96.2\% $$

Example 2: Working Backwards to Find Actual Yield

Sometimes, exam questions ask you to calculate the mass of product isolated if you know the efficiency of the reaction. For example, consider the reaction of $MnO_2$ with $HCl$:

$$ 4HCl + MnO_2 \rightarrow MnCl_2 + 2H_2O + Cl_2 $$

Suppose you calculate a theoretical yield of 20.96 g of $Cl_2$, but you know this reaction typically has a percent yield of 76.7%. What is the actual yield?

$$ \text{Actual Yield} = \text{Theoretical Yield} \times \frac{\text{Percent Yield}}{100} $$ $$ \text{Actual Yield} = 20.96 \text{ g} \times 0.767 = 16.08 \text{ g } Cl_2 $$

Advanced Concept: Atom Economy vs. Percent Yield

While percent yield tells you how much product you successfully isolated, it doesn't tell you how "green" or wasteless the reaction design is. This is where Atom Economy comes in.

Percent Yield: Efficiency of the process (human error, reaction completeness).
Atom Economy: Efficiency of the reaction design (how many atoms from reactants end up in the desired product vs. waste).

For example, in the thermite reaction ($Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$), if you are trying to produce Iron ($Fe$), the Aluminum Oxide ($Al_2O_3$) is a waste product. You could have a 100% yield (perfect conversion), but if the waste product is heavy, your atom economy might still be low. In modern chemistry, maximizing both yield and atom economy is the goal for sustainable science.

Summary of Stoichiometry Success

To consistently solve stoichiometry problems correctly, remember these rules:

Always Balance First: Never start calculations with an unbalanced equation.
Moles are the Bridge: You cannot convert grams of Reactant A to grams of Reactant B directly. You must go through moles using the balanced equation.
Trust the Limiting Reactant: The theoretical yield is always based on the reactant that runs out first.
Units Matter: Always keep track of grams and moles to avoid simple multiplication/division errors.