Limiting Reagent: Videos & Practice Problems

Understanding the limiting reagent is essential in stoichiometry, as it is the reactant completely consumed, determining the theoretical yield—the maximum product amount achievable. When multiple reactants have given amounts, stoichiometric calculations must be performed for each to identify the limiting and excess reagents. This process involves converting grams to moles, using balanced equation coefficients, and calculating moles of product. Mastery of these concepts enables accurate prediction of product yield and efficient chemical reaction analysis, reinforcing key principles like the law of conservation of mass and the role of reactants in chemical equations.

Limiting Reagent

In the study of stoichiometry, understanding the concept of the limiting reagent, also known as the limiting reactant, is crucial. The limiting reagent is the reactant that is entirely consumed during a chemical reaction, thereby determining the maximum amount of product that can be formed. This maximum amount is referred to as the theoretical yield, which represents the ideal or 100% yield of the reaction.

When multiple reactants are present in a chemical equation, identifying the limiting reagent becomes essential. The other reactants that remain after the reaction is complete are termed excess reagents. To ascertain which reactant is limiting and which is in excess, one must calculate the potential product yield from each reactant. This involves using stoichiometric relationships derived from the balanced chemical equation.

By analyzing the amounts of products that each reactant can produce, one can determine the limiting reagent and the excess reagent. This understanding is fundamental for predicting the outcomes of chemical reactions and optimizing reactant usage in practical applications.

Limiting Reagent

In stoichiometry, we often work with a single reactant to determine the amount of product formed in a chemical reaction. However, when dealing with multiple reactants, the process becomes more complex. Each reactant will have a specified amount, typically measured in grams, which must be converted to moles using the molar mass. This conversion is essential because stoichiometric calculations rely on the balanced chemical equation, where coefficients indicate the ratio of moles of reactants and products.For each reactant, the steps involve converting grams to moles, using the coefficients from the balanced equation to find moles of the desired product, and then converting those moles back into the desired units, such as grams, molecules, or ions. This dual approach allows us to calculate the potential yield of the product from each reactant.Through these calculations, we can identify the limiting reagent, which is the reactant that will be completely consumed first, thus determining the maximum amount of product that can be formed. The other reactant is termed the excess reagent, as it remains after the reaction has completed. The theoretical yield, which is the maximum amount of product expected from the reaction, can be calculated based on the limiting reagent.This methodical approach ensures that we accurately assess the quantities involved in reactions with multiple reactants, ultimately leading to a clearer understanding of the stoichiometric relationships at play.

Limiting Reagent Example

When chromium(III) oxide reacts with hydrogen sulfide gas, the reaction produces chromium(III) sulfide and water. The balanced chemical equation for this reaction is:

\(\mathrm{Cr_2O_3 (s) + 3H_2S (g) \rightarrow Cr_2S_3 (s) + 3H_2O (l)}\)

To determine the mass of chromium(III) sulfide formed when 14.2 grams of chromium(III) oxide reacts with 12.8 grams of hydrogen sulfide, the process begins by converting the given masses of reactants into moles. Using atomic masses from the periodic table—chromium (52 g/mol), oxygen (16 g/mol), hydrogen (1.008 g/mol), and sulfur (32.07 g/mol)—the molar mass of chromium(III) oxide (\(\mathrm{Cr_2O_3}\)) is calculated as:

\$2 \times 52 + 3 \times 16 = 152 \text{ g/mol}\(

Thus, moles of chromium(III) oxide are:

\)\frac{14.2 \text{ g}}{152 \text{ g/mol}} = 0.0934 \text{ mol}\(

Next, using the mole ratio from the balanced equation, 1 mole of \)\mathrm{Cr_2O_3}\( produces 1 mole of \)\mathrm{Cr_2S_3}\(. Therefore, moles of chromium(III) sulfide formed from chromium(III) oxide are also 0.0934 mol.

The molar mass of chromium(III) sulfide (\)\mathrm{Cr_2S_3}\() is calculated as:

\)2 \times 52 + 3 \times 32.07 = 200.21 \text{ g/mol}\(

Converting moles of \)\mathrm{Cr_2S_3}\( to grams:

\)0.0934 \text{ mol} \times 200.21 \text{ g/mol} = 18.7 \text{ g}\(

Similarly, for hydrogen sulfide (\)\mathrm{H_2S}\(), the molar mass is:

\)2 \times 1.008 + 32.07 = 34.086 \text{ g/mol}\(

Calculating moles of hydrogen sulfide:

\)\frac{12.8 \text{ g}}{34.086 \text{ g/mol}} = 0.3755 \text{ mol}\(

From the balanced equation, 3 moles of \)\mathrm{H_2S}\( produce 1 mole of \)\mathrm{Cr_2S_3}\(. Using this mole ratio:

\)0.3755 \text{ mol H}_2\mathrm{S} \times \frac{1 \text{ mol } \mathrm{Cr_2S_3}}{3 \text{ mol } \mathrm{H_2S}} = 0.1252 \text{ mol } \mathrm{Cr_2S_3}\(

Converting to grams:

\)0.1252 \text{ mol} \times 200.21 \text{ g/mol} = 25.06 \text{ g}$

Comparing the two calculated masses of chromium(III) sulfide, 18.7 g and 25.06 g, the smaller value represents the theoretical yield and corresponds to the limiting reagent, which in this case is chromium(III) oxide. The larger value corresponds to the excess reagent, hydrogen sulfide.

Therefore, the mass of chromium(III) sulfide formed is 18.7 grams, which is the theoretical yield based on the limiting reagent concept. This approach highlights the importance of identifying the limiting reagent to accurately predict product formation in chemical reactions.

Acrylonitrile (C­₃H₃N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction: 
2 C₃H₆ (g) + 2 NH₃ (g) + 3 O₂ (g)  → 2 C₃H₃N (g) + 6 H₂O (g)
If 12.0 g C₃H₆, 10.0 g NH­₃, and 5.0 g O₂ react, what mass of acrylonitrile can be produced, assuming 100% yield?

• A. 31.2 g
• B. 15.1 g
• C. 5.50 g
• D. 12.4 g
• E. 15.6 g

Answer: (C)

The reaction between solid aluminum and iron (III) oxide can generate temperatures reaching 3000 ºC and is used in welding metals. 
2 Al + Fe₂O₃ → Al₂O_­3 + 2 Fe
If 150 g of Al are reacted with 432 g of Fe­₂O₃, what is the mass of the excess reactant remaining? 

• A. 2.21 g
• B. 8.08 g
• C. 11.2 g
• D. 4.04 g
• E. 7.75 g

Answer: (D)

Here’s what students ask on this topic:

What is the limiting reagent in a chemical reaction and why is it important?

The limiting reagent, also called the limiting reactant, is the reactant that is completely consumed first in a chemical reaction. It determines the maximum amount of product that can be formed, known as the theoretical yield. This concept is important because it helps us predict how much product will be made before the reaction stops. Once the limiting reagent is used up, the reaction cannot proceed further, even if other reactants are still available. Understanding the limiting reagent allows chemists to optimize reactant amounts and calculate yields accurately.

How do you identify the limiting reagent when given amounts of multiple reactants?

To identify the limiting reagent when multiple reactants have given amounts, you perform stoichiometric calculations for each reactant separately. First, convert the given mass of each reactant to moles. Then, use the balanced chemical equation coefficients to calculate the moles of product each reactant can produce. The reactant that produces the least amount of product is the limiting reagent because it will be consumed first. The other reactant(s) are in excess and will remain after the reaction completes.

What is the difference between the limiting reagent and the excess reagent?

The limiting reagent is the reactant that is completely used up during a chemical reaction, limiting the amount of product formed. In contrast, the excess reagent is the reactant that remains after the reaction has finished because it was not fully consumed. The limiting reagent determines the theoretical yield, while the excess reagent is leftover material. Identifying both helps in calculating how much product forms and how much reactant remains unused.

How do you calculate the theoretical yield using the limiting reagent?

To calculate the theoretical yield, first identify the limiting reagent by comparing the amount of product each reactant can produce. Once the limiting reagent is found, use its mole amount and the balanced chemical equation to calculate the moles of product expected. Then convert these moles of product to grams or other desired units. This calculated amount is the theoretical yield, representing the maximum product possible if the reaction proceeds perfectly without any losses.

Why do stoichiometric calculations double when dealing with limiting reagents?

Stoichiometric calculations double when dealing with limiting reagents because you must perform the calculation for each reactant separately. Unlike simple stoichiometry where only one reactant amount is given, here you convert the mass of each reactant to moles, calculate the potential product from each, and then compare. This extra step is necessary to determine which reactant limits the reaction and to find the theoretical yield accurately.