Moles, excess and limiting reagents

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!

Recap of Molar Calculations

Molar calculations are fundamental in chemistry, providing a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure. These calculations are essential for stoichiometry, the quantitative study of chemical reactions. Let's explore the key concepts involved in molar calculations.

Understanding Moles

The mole is a unit of measurement in chemistry that represents a specific number of particles, typically atoms or molecules. One mole contains exactly 6.022 × 10^23 particles, known as Avogadro's number. This concept allows chemists to work with manageable quantities of substances rather than dealing with enormous numbers of individual particles.

Mass and Molar Mass

Mass is a fundamental property of matter, and in chemistry, we often work with molar mass. The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol). For example, the molar mass of carbon is 12.01 g/mol. This means that one mole of carbon atoms has a mass of 12.01 grams.

Molar Volume

Molar volume is the volume occupied by one mole of a substance. For gases at standard temperature and pressure (STP), the molar volume is approximately 22.4 liters per mole. This concept is particularly useful when dealing with gaseous substances in chemical reactions.

Concentration

Concentration refers to the amount of solute dissolved in a given amount of solvent. In molar calculations, we often use molarity, which is the number of moles of solute per liter of solution. For instance, a 1 M (molar) solution of NaCl contains one mole of NaCl per liter of solution.

Practical Examples of Molar Calculations

Let's consider some examples to illustrate these concepts:

1. Converting mass to moles: If you have 50 grams of NaCl (molar mass 58.44 g/mol), you can calculate the number of moles by dividing the mass by the molar mass: 50 g ÷ 58.44 g/mol = 0.856 moles of NaCl.
2. Molar volume calculation: At STP, 2 moles of any gas will occupy a volume of 2 × 22.4 L = 44.8 L.
3. Concentration calculation: To prepare 500 mL of a 0.5 M glucose solution, you would need 0.5 mol/L × 0.5 L = 0.25 moles of glucose.

Importance in Stoichiometry

Molar calculations are crucial in stoichiometry, which deals with the quantitative relationships between reactants and products in chemical reactions. By using moles, chemists can determine the amounts of substances needed for reactions or produced as a result of reactions.

For example, in the reaction 2H2 + O2 2H2O, we can deduce that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. This information allows us to calculate the quantities of reactants needed or products formed in real-world applications.

Conclusion

Mastering molar calculations is essential for anyone studying chemistry or working in related fields. These calculations provide a powerful tool for understanding and predicting chemical behavior, from simple lab experiments to complex industrial processes. By grasping the concepts of moles, mass, molar volume, and concentration, and applying them through stoichiometry, chemists can accurately analyze and control chemical reactions, paving the way for advancements in various scientific and technological domains.

Understanding Limiting Reagents

In chemistry, the concept of limiting reagents is crucial for understanding and predicting the outcomes of chemical reactions. A limiting reagent, also known as a limiting reactant, is the substance that determines the amount of product formed in a chemical reaction. It's the reactant that is completely consumed during the reaction, while the other reactants are in excess.

To grasp this concept better, let's consider a simple analogy. Imagine you're making sandwiches with 10 slices of bread and 4 slices of cheese. Each sandwich requires 2 slices of bread and 1 slice of cheese. In this case, the cheese is the limiting reagent because it will run out first, allowing you to make only 4 sandwiches, even though you have enough bread for 5.

Determining the limiting reagent is essential for calculating the theoretical yield of a reaction. Here's how to determine the limiting reactant in a chemical reaction:

How to Determine Limiting Reactant with Moles

1. Write a balanced chemical equation for the reaction.
2. Calculate the number of moles of each reactant present.
3. Divide the number of moles of each reactant by its coefficient in the balanced equation.
4. The reactant with the smallest resulting value is the limiting reagent.

For example, consider the reaction: 2H + O 2HO

If we have 6 moles of H and 4 moles of O:

• H: 6 moles ÷ 2 (coefficient) = 3
• O: 4 moles ÷ 1 (coefficient) = 4

In this case, H is the limiting reagent as it has the smaller value (3) after dividing by its coefficient.

How to Find Limiting Reactant with Grams

When working with grams instead of moles, the process is similar but requires an additional step:

1. Convert the mass of each reactant to moles using their molar masses.
2. Follow the steps for determining the limiting reagent with moles.

Let's use an example to illustrate this process:

Consider the reaction: 2Al + 3Cl 2AlCl

Given: 54 g of Al and 85 g of Cl

1. Convert to moles:
   • Al: 54 g ÷ 26.98 g/mol = 2 moles
   • Cl: 85 g ÷ 70.90 g/mol = 1.2 moles
2. Divide by coefficients:
   • Al: 2 moles ÷ 2 (coefficient) = 1
   • Cl: 1.2 moles ÷ 3 (coefficient) = 0.4

In this example, Cl is the limiting reagent as it has the smaller value (0.4) after dividing by its coefficient.

Limiting Reagent Calculation

Once you've identified the limiting reagent, you can use it to calculate the theoretical yield of the product. Here's a step-by-step guide for a limiting reagent calculation:

1. Identify the limiting reagent using the methods described above.

Excess Reagents: Concept and Calculations

In chemical reactions, understanding the concept of excess reagents is crucial for both theoretical and practical applications. An excess reactant, also known as an excess reagent, is the substance that remains partially unreacted when a chemical reaction is complete. This occurs when one reactant is present in a greater quantity than what is required to fully react with the other reactant(s). The importance of excess reagents lies in their ability to ensure complete consumption of the limiting reactant, maximize product yield, and drive reactions to completion.

To determine which reactant is in excess and calculate its quantity, chemists employ several methods. The first step is to identify the limiting reactant, which is the substance that will be completely consumed in the reaction. Once the limiting reactant is known, the excess reactant can be identified as the remaining reactant(s). To calculate the amount of excess reactant, follow these steps:

1. Write a balanced chemical equation for the reaction.
2. Determine the molar quantities of all reactants present.
3. Calculate the theoretical amount of product that could be formed from each reactant.
4. Identify the limiting reactant (the one that produces the least amount of product).
5. Calculate the amount of excess reactant that remains after the reaction is complete.

Let's illustrate this process with an example. Consider the reaction between hydrogen (H) and oxygen (O) to form water (HO):

2 H + O 2 HO

Suppose we have 10 moles of H and 7 moles of O. To determine the excess reactant:

1. The equation is already balanced.
2. We have 10 moles H and 7 moles O.
3. Theoretical product from H: 10 moles H ÷ 2 = 5 moles HO Theoretical product from O: 7 moles O × 2 = 14 moles HO
4. H is the limiting reactant as it produces less product.
5. O is in excess. Amount of excess O = 7 moles - (5 moles ÷ 2) = 4.5 moles

To find the mass of excess reactant, multiply the number of moles by the molar mass. In this case, if we want to know how to find the mass of excess reactant:

Mass of excess O = 4.5 moles × 32 g/mol = 144 g

Another method for excess reactant calculation involves using the concept of percent yield. This approach is particularly useful when dealing with experimental data. The steps for this method are:

1. Calculate the theoretical yield based on the limiting reactant.
2. Measure the actual yield of the product.
3. Calculate the percent yield: (Actual yield ÷ Theoretical yield) × 100
4. If the percent yield is less than 100%, the difference represents unreacted limiting reactant or side reactions.
5. Any remaining excess reactant can be calculated by subtracting the amount consumed from the initial amount.

Understanding how to calculate excess reactant and determine its mass is essential in various chemical processes, including industrial manufacturing, pharmaceutical synthesis, and academic research. Excess reagents play a vital role in:

• Ensuring complete reaction of the limiting reactant
• Shifting equilibrium to favor product formation
• Compensating for potential side reactions or impurities
• Improving reaction kinetics and yield
• Facilitating easier

  Practical Applications of Limiting and Excess Reagents

  Understanding limiting and excess reagents is crucial in chemistry and industry, with numerous practical applications that significantly impact various processes. The concept of limiting vs excess reactant plays a vital role in optimizing chemical reactions, improving efficiency, and reducing waste in both laboratory and industrial settings.

  One of the primary practical applications of limiting reagents is in the manufacturing sector. By identifying the limiting reagent, manufacturers can precisely calculate the maximum amount of product that can be formed in a reaction. This knowledge allows them to optimize raw material usage, reduce waste, and improve overall production efficiency. For instance, in the production of pharmaceuticals, where ingredients are often expensive and reactions must be carefully controlled, understanding limiting reagents helps ensure that resources are used effectively and product yield is maximized.

  In the field of environmental chemistry, the importance of identifying limiting reagents becomes evident when studying pollution control and remediation. For example, in wastewater treatment, understanding which chemical species is the limiting reagent can help engineers design more effective treatment processes. By focusing on the limiting factor, they can tailor treatment methods to address specific pollutants more efficiently, leading to cleaner water and reduced environmental impact.

  The concept of limiting reagents also finds applications in the food industry. In food processing and preservation, chemical reactions often play a crucial role. By identifying limiting reagents in these processes, food scientists can optimize ingredient ratios, improve product quality, and extend shelf life. This knowledge can lead to the development of more nutritious and longer-lasting food products, benefiting both consumers and producers.

  In the energy sector, particularly in the development of alternative fuels, understanding limiting reagents is essential. For instance, in the production of biodiesel, knowing the limiting reagent helps engineers design more efficient reactors and optimize the conversion process. This can lead to increased yield, reduced energy consumption, and ultimately, more sustainable fuel production.

  The importance of identifying limiting reagents extends to the field of materials science as well. In the synthesis of advanced materials, such as nanoparticles or specialized polymers, precise control over reaction conditions is critical. By identifying and manipulating limiting reagents, scientists can fine-tune material properties, achieve desired characteristics, and develop innovative materials for various applications, from electronics to medicine.

  In the pharmaceutical industry, optimizing chemical processes through the understanding of limiting reagents is crucial for drug development and production. This knowledge allows researchers to design more efficient synthesis routes, reduce the use of expensive or hazardous reagents, and improve overall drug yield. Additionally, it helps in scaling up reactions from laboratory to industrial production, ensuring consistent quality and efficacy of medications.

  The concept of limiting reagents also plays a significant role in analytical chemistry. In quantitative analysis, knowing the limiting reagent helps chemists accurately determine the concentration of unknown substances. This is particularly important in quality control processes, environmental monitoring, and forensic investigations, where precise measurements are essential.

  Furthermore, in the field of green chemistry, understanding limiting and excess reagents contributes to developing more sustainable chemical processes. By optimizing reactions based on limiting reagents, chemists can reduce the use of excess chemicals, minimize waste generation, and decrease the environmental footprint of chemical production.

  In conclusion, the practical applications of understanding limiting and excess reagents are vast and diverse. From improving manufacturing efficiency to advancing environmental protection, this knowledge is instrumental in optimizing chemical processes across various industries. By identifying limiting reagents, scientists and engineers can enhance product yield, reduce waste, improve resource utilization, and develop more sustainable practices. As chemistry continues to play a crucial role in addressing global challenges, the importance of mastering these fundamental concepts only grows, driving innovation and efficiency in countless fields.

  Problem-Solving Strategies for Limiting and Excess Reagent Calculations

  Welcome, chemistry enthusiasts! Today, we're diving into the fascinating world of limiting and excess reagents. Understanding these concepts is crucial for mastering stoichiometry and predicting the outcomes of chemical reactions. Let's break down the process step-by-step and explore some examples to help you become a pro at solving these problems.

  Step 1: Identify the Given Information

  First things first, carefully read the problem and identify the given information. This typically includes the balanced chemical equation and the amounts of reactants provided (usually in moles or grams).

  Step 2: Convert All Quantities to Moles

  To determine the limiting reagent, we need to work with moles. If the problem gives masses, convert them to moles using the molar mass of each substance. Remember, moles = mass / molar mass.

  Step 3: Calculate the Mole Ratios

  Use the balanced equation to determine the mole ratios of the reactants. This tells you how many moles of each reactant are needed for the reaction to occur completely.

  Step 4: Compare the Available Moles to the Required Moles

  Divide the number of moles available for each reactant by its coefficient in the balanced equation. The reactant that yields the smallest value is the limiting reagent.

  Step 5: Calculate the Amount of Product Formed

  Use the limiting reagent to determine the amount of product that can be formed. This involves using the mole ratio between the limiting reagent and the product.

  Step 6: Determine the Amount of Excess Reagent Remaining

  Calculate how much of the excess reagent remains after the reaction is complete. Subtract the amount used from the initial amount provided.

  Example 1: A Simple Limiting Reagent Problem

  Let's start with a basic example. Consider the reaction: 2H + O 2HO
  Given: 4 moles of H and 3 moles of O
  Step 1: Information is already in moles.
  Step 2: Not needed here.
  Step 3: The ratio is 2:1 (H:O)
  Step 4: H: 4/2 = 2, O: 3/1 = 3. O yields the smaller value, so it's the limiting reagent.
  Step 5: 3 moles of O will produce 2 × 3 = 6 moles of HO.
  Step 6: Excess H = 4 - (2 × 3) = 1 mole

  Example 2: Limiting Reagent with Mass Conversion

  Now, let's try a more complex problem. Consider the reaction: 2Al + 3Cl 2AlCl
  Given: 54 g of Al and 106.5 g of Cl
  Step 1: Identify given information.
  Step 2: Convert to moles:
  Al: 54 g / (27 g/mol) = 2 moles
  Cl: 106.5 g / (71 g/mol) = 1.5 moles
  Step 3: The ratio is 2:3 (Al:Cl)
  Step 4: Al: 2/2 = 1, Cl: 1.5/3 = 0.5. Cl yields the smaller value, so it's the limiting reagent.
  Step 5: 1.5 m

  Conclusion: Mastering Moles, Excess and Limiting Reagents

  In this article, we've explored the fundamental concepts of moles in stoichiometry, excess reagents, and limiting reagents in stoichiometry. Understanding these principles is crucial for accurate chemical calculations and predicting reaction outcomes. We've learned how to identify limiting reagents, calculate excess amounts, and determine theoretical yields. These concepts form the backbone of quantitative chemistry and are essential for both academic and practical applications. To truly master these topics, regular practice and problem-solving are key. Revisit the introduction video for a visual reinforcement of these concepts, as it provides a solid foundation for understanding stoichiometric relationships. By grasping these principles, you'll be better equipped to tackle more complex chemical problems and deepen your understanding of reaction dynamics. Remember, proficiency in moles in stoichiometry and reagent calculations is a valuable skill that will serve you well throughout your chemistry journey.

  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 $6.022 \times 10^{23}$ 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.

  Prerequisite Topics

  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.

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.

$Mol \, HCl = 750 \, mL \, HCl \,$ $\large \frac{1 \, L \, HCl} {1000 \, mL \, HCl} \, * \, \frac{0.5 \, mol \, HCl } {1 \, L \, HCl }$ = 0.375 $mol \, HCl$

$Mol \, NaOH = 625 \, mL \, NaOH \,$ $\large \frac{1 \, L \, NaOH} {1000 \, mL \, NaOH} \, * \, \frac{0.8 \, mol \, NaOH } {1 \, L \, NaOH }$ = 0.5 $mol \, NaOH$
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$\, mol \, NaOH -$ 0.375 $mol \, HCl$ = 0.125 $\, mol \, NaOH \,$ 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$\, mol \, NaOH \, * \,$ $\large \frac{1 \, L \, NaOH} {0.8 \, mol \, NaOH} \, * \, \frac{1000 \, mL \, NaOH } {1 \, L \, NaOH }$ = 156.25 $\, mL \, NaOH$
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.