There are many things that need to go right for a chemical reaction to yield useful products: from the environment surrounding the reaction to the amount of the reactants present. Only once in a blue moon do all the reactants get converted into products. In most reactions, one reagent (reagent and reactant are used interchangeably) is entirely depleted, while some quantity of the other reagents stays available for further reaction. Since one of the reactants is not always available, the reaction hits a roadblock and does not continue. This reactant that gets completely used up, and thus limits the reaction from advancing forward, is called the limiting reactant or limiting reagent.
The limiting reagent in a chemical reaction controls how much of the final product will be produced. Recommended Video for you: What is limiting reactant?Based on their amounts and their roles, we classify reactants into two kinds: limiting reactants and excess reactants. Limiting reactants are those that get completely utilized in a reaction first and thus limit the amount of product that will be produced. Excess reactants, on the other hand, are the reactants that are still present after the reaction has reached a standstill. Let’s say that you’re standing in a queue at your favorite bagel vendor. The bagel guy comes out of the cart and announces that he only has 10 bagels left. Now, you look up from your phone and start counting the number of people ahead of you in the queue. You count a total of 20 people, which is 10 more than the number of bagels that remain. In our example, the customers standing in the queue and the bagels are reactants, coming together to produce happy-fed individuals (end product). However, the number of bagels will limit the number of happy-fed customers that can be achieved in the end, making it the limiting factor (agent), while the customers will be considered the excess reactants. Another example is the number of buns and sausages needed for making a hot dog. It takes both a bun and a sausage to make a hot dog. An unequal number of the items (reactants) will result in an inadequate number of hot dogs (product). With 10 buns and 5 sausages, we will only be able to produce 5 hot dogs and will have 5 buns remaining. The buns are available in excess, while sausages are the limiting agent, controlling the number of hot dogs that can be made. How to find the limiting reactant?There are a couple of ways by which one can determine the limiting and excess reactants in a reaction. However, the methods come with a prerequisite, which is to have a balanced chemical equation in hand. The methods use stoichiometric coefficients from the balanced chemical equation to calculate ratios; if the coefficients themselves are incorrect, the final answers obtained will also be incorrect. In the first method, we will find and compare the mole ratios of the reactants, while in the other one, we will find the amount of product that will be produced by each reactant. The one that produces the least amount of the end product is the limiting reagent. Method 1: Using Mole RatiosLet’s apply this method to the reaction of ammonia (NH3) and molecular oxygen (O2) to figure out the limiting reactant of the two. The reaction between NH3 and O2 yields NO (nitric oxide) and H2O (water). The balanced chemical equation for the reaction is: 4NH3 + 5O2 → 4NO + 6H2O The equation is read as 4 moles of ammonia require 5 moles of oxygen to produce 4 moles of nitric acid and 6 moles of water. We start by finding the stoichiometric ratio of the reactants. To do so, we simply divide the stoichiometric coefficients of each reactant. For ammonia, the stoichiometric coefficient is 4, while for oxygen it is 5. Stoichiometric Ratio = 4 moles of ammonia / 5 moles of oxygen = 0.8 moles of ammonia/1 mole of oxygen This implies that we need 0.8 moles of ammonia for each mole of oxygen. Next, we find the mole ratios of the reactants. In most scenarios, the quantities of the reactants will be provided in grams. We convert the gram values to moles and then continue towards finding the limiting reactant. Let’s assume that we have 100 grams of ammonia and oxygen each. To convert this into moles, we divide these values by their molecular masses. For ammonia, the number of moles = (100 g) / (17.04 g/mol) = 5.86 mol For oxygen, the number of moles = (100 g) / (32 g/mol) = 3.125 mol We can continue forward in two ways. One, by assuming a limiting reactant and finding the number of moles of other reactants required, or the other, by finding the actual ratio and drawing a conclusion from its value. Let’s go down the first path and assume ammonia to be the limiting reactant. Assuming ammonia is consumed first, (5.86 moles of ammonia) / Stoichiometric Ratio = 4.688 moles of oxygen are required. But we only have 3.125 moles of oxygen available for the reaction, so we will run out of oxygen before ammonia. Therefore, oxygen is the limiting reactant and ammonia is available in excess. Method 2: By comparing the amount of product produced by each reactantJust as in the previous method, we start off with a balanced chemical equation and continue by determining the number of moles of each reactant. We will make use of the same values as those found in the previous method. The number of moles of ammonia = 5.86 mol & the number of moles of oxygen = 3.125 mol. Now we find the amount of the end product (NO) that each reactant will produce. For ammonia, the mol of nitric oxide (NO) = Moles of NH3 available × Stoichiometric coefficient of NO/ stoichiometric coefficient of NH3 = 5.86 × 4/4 = 5.86 mol of NO For oxygen, the mol of nitric oxide (NO) = Moles of O2 available × Stoichiometric coefficient of NO/ stoichiometric coefficient of O2 = 3.125 × 4/5 = 2.5 mol of NO Since the amount of product produced by oxygen is less than that produced by ammonia, oxygen is the limiting reactant and ammonia is in excess. Final wordsAs we can see, the limiting reagent or limiting reactant in a reaction is the reactant that gets completely exhausted and thus prevents the reaction from continuing forward. It also determines the amount of the final product that will be produced. Finding the limiting reactant is an important step in finding the percentage yield of the reaction. The percentage yield of a reaction is the ratio of its actual yield to its theoretical yield times 100. Theoretical yield is the yield predicted by stoichiometric calculations, assuming the limiting reactant reacts completely. In simpler words, it is the amount of product produced from the limiting reactant. In our case, the limiting reactant is oxygen and the amount of product (NO) produced from it is 2.5 moles. Thus, the theoretical yield for the reaction is 2.5 moles. The actual yield is the amount of end product obtained upon experimentation. Let’s assume the actual yield we obtained on experimentation as 2 moles. The percentage yield will then be: Percentage yield = (Actual yield/Theoretical yield) × 100 = (2/2.5) × 100 = 80% The theoretical yield calculated is usually higher than the actual yield that is produced. This happens due to a variety of reasons, including the reversible nature of reactions, the production of undesired by-products, errors in purification processes, etc. Suggested ReadingWas this article helpful?
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Limiting reactant Definition
This reactant is much important while dealing with a chemical reaction. The reactant that controls the amount of products formed in a chemical reaction and is consumed earlier is called a limiting reactant. Watch the video lecture to better understand about limiting reactant Video lecture of Limiting reactantExplanation of limiting reactantIf stoichiometry amounts are put in chemical reactions, these amounts are completely used. Sometimes, stoichiometric amounts required by the reaction are not used during the reaction for the following purposes:
When we use non-stoichiometric amounts, one of the reactants is consumed earlier than the other. The reactant which is used in less amount is of products formed and is called a limiting reactant. Related: Avogadro’s number and Molar Volume-Themasterchemistry.com How to find limiting reactantTo determine the limiting reactant following steps should be taken
Examples to understand a limiting reactantExample 1 In a chemical reaction, a large quantity of oxygen makes things burn rapidly. Since oxygen used in excess is left behind when the reaction is completed. The other reagent is consumed completely. That reactant that is consumed earlier is known as limiting reactant. Example 2 The concept of limiting reactant is quite similar to the relation between 30 eggs and 58 slices to make the sandwiches. As 1 egg and 2 slices make 1 sandwich. Therefore, only 29 sandwiches can be prepared from 58 slices. One egg is left behind. 58 slices are completely used which limits the product sandwiches. So slices are limiting reactants. Example 3 Consider the reaction along with stoichiometric amounts 2H2 + O2 ———-> 2H2O (2 moles) (1 mole) (2 moles) 4g 32g 36g If we use stoichiometric amounts then 4g amounts of hydrogen and 32 g of oxygen, we get 36g of water. But if we take non-stoichiometric amounts say 4g of H2 and 60g of oxygen, we will not get more quantity of water than 36g because H2 is less. So H2 is a limiting reactant. 28 g of oxygen will be left unreacted. 2H2 + O2 ——-> 2H2O 4g 36 g 36g Limiting reactant problemsProblem 1 NH3 gas can be prepared by heating together two solids, NH4Cl and Ca(OH)2. If a mixture containing 100 g of each solid is heated, then a) Calculate the number of grams of NH3 produced b) Calculate the excess amount of reagent left unreacted Solution The balanced chemical equation for the chemical reaction is 2NH4Cl + Ca(OH)2 —-> CaCl2 + 2NH3 + 2H2O a) Number of grams of NH3 produced given mass of NH4Cl = 100g Molar mass of NH4Cl = 14+1×4+35.5 = 53.5 g/mol Number of moles of NH4Cl= 100/53.5 = 1.87 moles Given mass of Ca(OH)2 = 100 g Molar mass of Ca(OH)2 = 40 + 2(16+1) = 74 g Number of moles of Ca(OH)2 = 100/74 = 1.35 moles –> Compare NH4Cl and NH3 according to a balanced chemical equation. NH4Cl : NH3 2 moles : 2 moles Therefore 1.87 moles: 2/2 x 1.87 = 1.87 moles –> Compare Ca(OH)2 and NH3 according to the balanced chemical equation Ca(OH)2 : NH3 2 moles : 2 moles Therefore, 1.35 moles: 2 x 1.35= 2.70 moles Since NH4Cl produces the least amount of NH3, hence NH4Cl is the limiting reactant. Thus no. of moles of NH3 produced= 1.87 moles Molar mass of NH3 = 14+1 x3 = 17 g/mol Hence the amount of NH3 produced = No.of moles x Molar mass of NH3 as given below. Amount of NH3 produced = 1.87 x 17 = 31.79 g b) Amount of reagent left unreacted –> Compare NH4Cl and Ca(OH)2 according to a balanced chemical equation. NH4Cl : Ca(OH)2 2 moles : 1 mole 1.87 moles : 1×1.87/2 = 0.935 moles Moles of Ca(OH)2 taken = 1.35 moles Therefore unreacted moles = 1.35 – 0.935 = 0.415 moles Thus mass of Ca(OH)2 left = 0.415 x 74 = 30.71 g Problem 2 Calculate the number of grams of Al2S3, which can be prepared by the reaction of 20 g of Al and 20g of sulfur. How much non-limiting reactant is in excess? Solution The balanced chemical equation for the chemical reaction is 2Al + 3S —–> Al2S3 Given mass of Al = 20 g Number of moles of Al = 20/27 = 0.74 moles Given mass of S = 30 g Number of moles of S = 30/32 = 0.9375 moles Determination of limiting reactant in this reaction Compare Al and Al2S3 according to a balanced chemical equation. Al : Al2S3 2 moles : 1 mole Therefore, 0.74 moles: ½ x 0.74 = 0.37 moles Compare the number of moles of the product produced by Al = 0.37 –> Compare S and Al2S3 according to a balanced chemical equation. S : Al2S3 3 moles : 1 mole Therefore, 0.9375 moles: 1/3 x 0.9375 = 0.3215 moles Hence the number of moles of Al2S3 produced by S = 0.3225 moles Since S produces the least number of moles of the product, therefore it is the limiting reactant. Hence moles of Al2S3 produced = 0.3125 moles Mass of Al2S3 produced = 0.3125 x 150 = 46.87 g Determination of amount of Al left unreacted 2Al + 3S ——> Al2S3 –> Compare the moles of S and Al to find the moles of Al reacted S : Al 3 moles : 2 moles 0.9375 moles: 2/3 x 0.9375 = 0.625 moles Moles of Al consumed = 0.625 moles Moles of Al taken = 0.74 moles Moles of Al left unreacted = 0.74 – 0.625 = 0.115 moles Mass of Al left unreacted: 0.115 x 27 = 3.105 g Related questions1: How does a limiting reactant help to control the reaction?As we know a limiting reactant is the one that is in a lower amount and is consumed earlier. After complete consumption of limiting reactant further formation of product stops and excess reagent is left in the reaction media. Therefore, if the limiting reagent is not available to the excess reagent then the product cannot be formed further. In this way, a limiting reagent controls the chemical reaction. 2: 11g of carbon is reacted with 32 g of oxygen to give CO2. Which is the limiting reactant?C + O2 ——> CO2 According to a balanced chemical equation, there should be 44 grams of CO2 after the complete reaction of 12g of carbon with 32 grams of oxygen. Therefore, 32 grams of oxygen will be in excess compared to 11 grams of carbon. So in this case carbon is a limiting reactant. 3: Why the concept of limiting reactants is not applicable to reversible reactions?For a reaction having one of the reactants as a limiting one, one of the reactants has to be consumed completely. But in the case of a reversible reaction, at the stage of equilibrium, certain amounts of reactants are left behind. Therefore, the reversible reactions do not give any idea about the concept of limiting reactants. 4. How do you find the limiting reactant?You need to perform a titration analysis to determine the limiting reactant. For example, if you have two reactants, A and B, then you can determine the limiting reactant by using a stoichiometric equation, such as (A + B) = C. 5. What is a limiting and excess reactant?Reactants are substances that react to form new compounds. Limiting reactants are the ones that cannot be combined to form a compound with a greater quantity of the element. Excess reactants are those that are left after you’ve added the limiting reactants to your reaction. They can be used again in future reactions. 6. What is an excess reactant?An excess reactant is a substance that is left over after the reaction of a chemical equation has taken place. An example of this would be when you mix a cup of salt and sugar to create a rock candy, and you end up with about 2 cups of salt and 1/2 cup of sugar. In this case, the remaining 1/2 cup of sugar is the excess reactant. |
Which of the following methods can be used to determine the limiting reagent in an equation
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