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Mastering Moles, Excess and Limiting Reagents
Dive into the world of stoichiometry! Learn to identify limiting reagents, calculate excess reactants, and predict reaction outcomes. Perfect for students aiming to excel in chemistry and real-world applications.

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Now Playing:Stoichiometry of limiting and excess reagents – Example 0a
Intros
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  1. Recap: Moles and chemical reactions.
  2. The limiting reagents.
Examples
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  1. Find the limiting reagent in a chemical reaction with known quantities.
    Consider the reaction:

    2H2+_2 + O2_2 →2H2_2 O

    1. 50 g of O2_2 gas and 50 g of H2_2 gas were reacted together.
      Which reagent is the limiting reagent?

    2. What mass of the other reagent is in excess?

Introduction to stoichiometry
Jump to:NotesConceptExampleFAQsPrerequisites
Notes
In this lesson, we will learn:
  • The importance of identifying the limiting reagent in reactions.
  • To identify by calculation the limiting and excess reagents in a chemical reaction.
  • To calculate quantities in excess.

Notes:

  • In the past few lessons we have learned to calculate amounts of substance used in reactions for the solid, gas and aqueous phases.
    Moles / stoichiometry test questions usually involve being given one quantity of a reactant or product and:
    • Converting the quantity of it from one unit into another unit;
    • Converting this unit into moles;
    • Using the molar or stoichiometric ratio of the reaction to find the moles of another substance;
    • Converting the moles of that new substance into another quantity for your final answer.
    You might only be asked about one reactant in a reaction, but remember for a reaction to go, every reactant in the equation must be present.

  • For a chemical reaction to happen, all the reactants must be present and available to react. If even one reactant is not present, the reaction will not happen; everything must be there. In real chemical reactions, this means that a reaction will go until one of the reactants has run out. When this happens, the reaction stops.
    • The chemical that you have the least amount of, or that runs out first is called the limiting reagent because its running out limits the reaction from happening any longer.
    • Reagents that are not limiting reagents are excess reagents or are “in excess”. We call it this because when the limiting reagent runs out, there will still be some of this reagent left over – an ‘excess’ amount of it.
    This is important to chemists as they plan reactions because we are doing them to obtain the products. If we only have x moles of a reactant, we can only expect y moles of product.

  • To find out the limiting reagent, you need to find the amount of product that can be made, with respect to each reactant involved. The reactant that would produce the smallest amount of product is the limiting reagent.
    To find the mass of excess reagent, find the amount of the excess reagent that reacts based on the amount of limiting reagent. Then, subtract that from the total amount of excess reagent available.
    Chemists do chemical reactions because we want the valuable products of them. This is why we find limiting reagents in terms of the amount of products we can get. This is also why finding excess reagent quantities is important; this excess is going to do nothing in the reaction unless we have more limiting reagent.

  • Worked example: Find the limiting reagent and quantity of excess.

    • Consider the reaction:

    HCl + NaOH   \, →  \, NaCl + H2O

    Two aqueous solutions, one of 0.5 M HCl and 0.8 M NaOH can be mixed together to produce NaCl. 750 mL of the HCl solution is available, while 625 mL of the NaOH solution is available.

    Identify the limiting reagent in this reactant, and the quantity of excess reagent in mL.

    The first step in this problem is to find the number of moles of both reagents. Both are required, and one will run out before the other, so we need to calculate how much of both we have. The reagent with less moles is the limiting reagent.

    Mol HCl=750 mL HCl Mol \, HCl = 750 \, mL \, HCl \, 1 L HCl1000 mL HCl ∗ 0.5 mol HCl1 L HCl\large \frac{1 \, L \, HCl} {1000 \, mL \, HCl} \, * \, \frac{0.5 \, mol \, HCl } {1 \, L \, HCl } = 0.375 mol HClmol \, HCl


    Mol NaOH=625 mL NaOH Mol \, NaOH = 625 \, mL \, NaOH \, 1 L NaOH1000 mL NaOH ∗ 0.8 mol NaOH1 L NaOH\large \frac{1 \, L \, NaOH} {1000 \, mL \, NaOH} \, * \, \frac{0.8 \, mol \, NaOH } {1 \, L \, NaOH } = 0.5 mol NaOHmol \, NaOH

    From these calculations we can see that the NaOH is in excess, so HCl is the limiting reagent. How much of the NaOH is in excess?

    0.5 mol NaOH−\, mol \, NaOH - 0.375 mol HClmol \, HCl = 0.125  mol NaOH \, mol \, NaOH \, excess reagent


    Now we know that 0.125 moles of NaOH is in excess. This will not react because it doesn’t have any HCl to react with. How much volume of our NaOH solution contains 0.125 moles of NaOH, which we don’t need to use in the reaction?

    0.125 mol NaOH ∗ \, mol \, NaOH \, * \, 1 L NaOH0.8 mol NaOH ∗ 1000 mL NaOH1 L NaOH\large \frac{1 \, L \, NaOH} {0.8 \, mol \, NaOH} \, * \, \frac{1000 \, mL \, NaOH } {1 \, L \, NaOH } = 156.25  mL NaOH\, mL \, NaOH

    So this reaction will make as much product as possible with the entire solution of HCl, and 156.25 mL of the NaOH solution can be spared as this is in excess. This assumes the reaction goes to completion.

    Knowing your limiting reagent and the amount you have is important because a limit of reagents available puts a limit on the amount of products you can make too! Calculating the limiting reagent and the quantity of it manages expectations of our product yield, which we will learn about next lesson in Percentage yield and atom economy.
Concept

Introduction: Understanding Moles, Excess and Limiting Reagents

Welcome to our exploration of moles, excess reagents, and limiting reagents! These concepts are fundamental to stoichiometry and crucial for understanding chemical reactions. Our introduction video will guide you through these topics, providing a solid foundation for your chemistry journey. Moles are the basic unit for measuring substances in chemistry, allowing us to quantify particles on an atomic scale. When we dive into chemical reactions, we'll encounter excess reagents - those present in quantities greater than necessary for the reaction. On the flip side, limiting reagents are the substances that determine the amount of product formed. Understanding these concepts is key to predicting reaction outcomes and optimizing chemical processes. As we progress, you'll see how these ideas apply to real-world scenarios, from industrial manufacturing to environmental science. So, let's embark on this exciting journey into the world of chemical stoichiometry together!

Example

Recap: Moles and chemical reactions.

Step 1: Understanding Moles

Moles are a fundamental concept in chemistry that represent an amount of chemical substance. One mole is defined as 6.022×10236.022 \times 10^{23} entities (Avogadro's number) of a given substance. This can be atoms, molecules, ions, or other particles. The concept of moles allows chemists to count particles by weighing them. For example, one mole of carbon atoms weighs 12 grams because the atomic mass of carbon is 12 atomic mass units (amu).

Step 2: Calculating Moles from Mass

To calculate the number of moles from a given mass, you need to know the molar mass of the substance. The molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). The formula to calculate moles from mass is:
Number of moles = Mass (g) / Molar mass (g/mol)
For example, if you have 24 grams of carbon, the number of moles of carbon would be:
Number of moles = 24 g / 12 g/mol = 2 moles

Step 3: Understanding Limiting Reagents

In a chemical reaction, the limiting reagent is the reactant that is completely consumed first, limiting the amount of product that can be formed. The other reactants that are not completely used up are called excess reagents. Identifying the limiting reagent is crucial for calculating the theoretical yield of a reaction.

Step 4: Calculating Limiting Reagents

To determine the limiting reagent, follow these steps:

  1. Write the balanced chemical equation for the reaction.
  2. Calculate the number of moles of each reactant.
  3. Use the stoichiometry of the balanced equation to determine the mole ratio between the reactants and the products.
  4. Compare the mole ratio of the reactants to identify which one is the limiting reagent.
For example, consider the reaction between hydrogen and oxygen to form water:
2H2 + O2 2H2O
If you have 4 moles of H2 and 1 mole of O2, the mole ratio of H2 to O2 is 4:1. According to the balanced equation, the required ratio is 2:1. Since you have more than enough H2 to react with the available O2, O2 is the limiting reagent.

Step 5: Calculating Excess Reagents

Once the limiting reagent is identified, you can calculate the amount of excess reagent left over after the reaction. Use the stoichiometry of the balanced equation to determine how much of the excess reagent reacts with the limiting reagent. Subtract this amount from the initial amount of the excess reagent to find the remaining quantity.
For example, in the reaction above, if you start with 4 moles of H2 and 1 mole of O2, and O2 is the limiting reagent, you can calculate the amount of H2 that reacts:
2 moles of H2 react with 1 mole of O2
Since you have 4 moles of H2, only 2 moles will react with the 1 mole of O2. The remaining H2 will be:
4 moles - 2 moles = 2 moles of H2 left over

Step 6: Recap and Application

Understanding moles, limiting reagents, and excess reagents is essential for predicting the outcomes of chemical reactions and for practical applications in laboratory settings. By mastering these concepts, you can accurately determine the quantities of reactants and products involved in a reaction, optimize the use of materials, and minimize waste.

FAQs

Here are some frequently asked questions about moles, excess and limiting reagents:

1. What is a limiting reagent?

A limiting reagent is the reactant in a chemical reaction that is completely consumed and determines the amount of product that can be formed. It limits the extent of the reaction and is used up first.

2. How do you identify the limiting reagent?

To identify the limiting reagent: 1. Balance the chemical equation. 2. Calculate the moles of each reactant. 3. Divide the moles of each reactant by its coefficient in the balanced equation. 4. The reactant with the smallest resulting value is the limiting reagent.

3. What is an excess reagent?

An excess reagent is a reactant present in a quantity greater than necessary for the reaction to go to completion. It remains partially unconsumed after the limiting reagent is used up.

4. How do you calculate the amount of excess reagent remaining?

To calculate the amount of excess reagent remaining: 1. Determine the limiting reagent. 2. Calculate how much of the excess reagent is consumed based on the limiting reagent. 3. Subtract the amount consumed from the initial amount of excess reagent.

5. Why are moles important in stoichiometry calculations?

Moles are crucial in stoichiometry because they allow chemists to relate the number of particles at the atomic level to measurable quantities in the lab. They provide a bridge between the microscopic and macroscopic worlds, enabling accurate predictions and calculations in chemical reactions.

Prerequisites

Understanding the concept of moles, excess reagents, and limiting reagents is crucial in chemistry, but to truly grasp these ideas, it's essential to have a solid foundation in prerequisite topics. One of the most important prerequisites for this subject is balancing chemical equations. This fundamental skill is not just a stepping stone; it's an integral part of working with moles and reagents in chemical reactions.

When dealing with moles, excess, and limiting reagents, you're essentially analyzing the quantities of substances in a chemical reaction. However, before you can even begin to consider these quantities, you need to ensure that the chemical equation itself is correctly balanced. Balanced chemical equations provide the foundation for all subsequent calculations and analyses in this area.

Consider this: when you're trying to determine which reagent is limiting in a reaction, you're comparing the relative amounts of reactants based on the balanced equation. Without a properly balanced equation, it's impossible to accurately assess which substance will be completely consumed first. Similarly, identifying excess reagents relies on understanding the stoichiometric ratios established by a balanced equation.

Moreover, the concept of moles is intrinsically linked to balanced equations. The coefficients in a balanced chemical equation represent the relative number of moles of each substance involved in the reaction. This relationship is fundamental when calculating the amounts of products formed or reactants consumed in a chemical process.

By mastering the art of balancing chemical equations, you're not just learning a isolated skill; you're building a critical foundation for understanding more complex concepts like moles, excess reagents, and limiting reagents. This prerequisite knowledge allows you to visualize the proportions of substances in a reaction, making it easier to grasp how different quantities of reactants interact and influence the outcome of a chemical process.

In conclusion, while focusing on moles, excess, and limiting reagents, never underestimate the importance of prerequisites like balancing chemical equations. This fundamental skill is the key to unlocking a deeper understanding of stoichiometry and reaction analysis. By solidifying your knowledge of these basics, you'll find yourself better equipped to tackle more advanced chemical concepts with confidence and clarity.