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Stoichiometry III Section 12.3: Limiting Reagent and Percent Yield.

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Presentation on theme: "Stoichiometry III Section 12.3: Limiting Reagent and Percent Yield."— Presentation transcript:

1 Stoichiometry III Section 12.3: Limiting Reagent and Percent Yield

2 Objectives Upon completion of this presentation, you will be able to identify the limiting reagent in a reaction. calculate the theoretical yield, actual yield, or percent yield of a chemical reaction given appropriate information.

3 Introduction Let’s say that we are going to open up a sandwich shop. Eventually, we are going to make a variety of sandwiches to sell. But now, we are only going to offer one sandwich: a roast beef sandwich. A single roast beef sandwich is made up of – ¼ lb. of beef 2 slices of bread 1 leaf of lettuce ½ tomatoes, sliced 2 Tsp. of mayonnaise 1 Tsp. of mustard

4 The next thing we need to do is to get our supplies. We buy the following: 50 lb of roast beef 100 loaves of bread 10 heads of lettuce 50 tomatoes 10–16 oz. jars of mayonnaise 5–8 oz. jars of mustard

5 How many sandwiches can we make? We can make 200 sandwiches from 50 lb of roast beef (50 lb./¼ lb. per sandwich = 200 sandwiches). We can make 1,000 sandwiches from 100 loaves of bread (each loaf has 20 slices; that’s 10 sandwiches per loaf; 10×100 = 1,000 sandwiches). We can make 120 sandwiches from 10 heads of lettuce (each head has about 12 good leaves; 12×10 = 120 sandwiches). We can make 100 sandwiches from 50 tomatoes (we use ½ a tomato per sandwich; 2×50 = 100 sandwiches). We can make 160 sandwiches from 10–16 oz. jars of mayonnaise (2 Tsp. is 1 oz.; 10×16 = 160 sandwiches). We can make 80 sandwiches from 5–8 oz. jars of mustard (2 Tsp. is 1 oz.; 2×8×5 = 80 sandwiches).

6 Let’s look at this in another form. The maximum number of sandwiches we can make is 80 sandwiches. The amount of mustard limits the number of sandwiches we can make. In chemistry, we call this the limiting reagent. ingredientquantitysandwiches Roast beef50 lb.200 Bread100 loaves1,000 Lettuce10 heads120 Tomatoes50 tomatoes100 Mayonnaise10–16 oz. jars160 Mustard5–8 oz. jars80

7 Limiting and Excess Reagents In any chemical reaction, the amount of product is limited by the amount of reactants. The more reactants we have, the more product we can get. When we have a reaction with more than one reactant, the reactant that runs out first is called the limiting reagent. The reactant that does not run out is call the excess reagent.

8 Limiting and Excess Reagents In order to determine the amount of product we should get from a reaction, we need to know the limiting reagent. We use stoichiometry to do this.

9 Limiting and Excess Reagents For example, 48.0 g of methane, CH 4, is burned with 160. g of oxygen gas. What is the limiting reagent (which will run out faster, CH 4 or O 2 )? CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) M (g/mol) m (g) n (mol) ____ ____ ①② First, we calculate the number of mols of each reactant. ① n CH4 = m CH4 / M CH4 = (48.0 g)/(16.0 g/mol) = 3.00 mol 3.00 ② n O2 = m O2 / M O2 = (160. g)/(32.0 g/mol) = 5.00 mol 5.00

10 Limiting and Excess Reagents For example, 48.0 g of methane, CH 4, is burned with 160. g of oxygen gas. What is the limiting reagent (which will run out faster, CH 4 or O 2 )? CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) M (g/mol) m (g) n (mol) Next, we calculate the number of mols of O 2 needed to react with all of the CH n CH4 n O2 = 1 2 ⇒ 1 × n O2 = 2 × n CH4 ⇒ n O2 = 2 × 3.00 mol = 6.00 mol But, we only have 5.00 mols of O 2. We will run out of O 2 before we run out of CH 4. O 2 is the limiting reagent.

11 Limiting and Excess Reagents For example, 56.0 g of nitrogen reacts with 18.0 g of hydrogen to form ammonia. What is the limiting reagent (which will run out faster, N 2 or H 2 )? N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) M (g/mol) m (g) n (mol) ____ ____ ①② First, we calculate the number of mols of each reactant. ① n N2 = m N2 / M N2 = (56.0g)/(28.0 g/mol) = 2.00 mol 2.00 ② n H2 = m H2 / M H2 = (18.0 g)/(2.02 g/mol) = 9.00 mol 9.00

12 Limiting and Excess Reagents For example, 56.0 g of nitrogen reacts with 18.0 g of hydrogen to form ammonia. What is the limiting reagent (which will run out faster, N 2 or H 2 )? N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) M (g/mol) m (g) n (mol) ____ ____ ①② Next, we calculate the number of mols of H 2 needed to react with all of the N 2. n N2 n H2 = 1 3 ⇒ 1 × n H2 = 3 × n N2 ⇒ n H2 = 3 × 2.00 mol = 6.00 mol But, we have 9.00 mols of H 2. We will run out of N 2 before we run out of H 2. N 2 is the limiting reagent.

13 Limiting and Excess Reagents Sample problems. Find the limiting reagent for each of the following reactions. 1. N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) [28.0 g N 2 & 2.50 g H 2 ] 2. CaO( s ) + CO 2 ( g ) → CaCO 3 ( s ) [112 g CaO & 66.0 g CO 2 ] 3. 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O( l ) [12.0 g H 2 & 16.0 g O 2 ] 4. CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) [32.0 g CH 4 & 168 g O 2 ] 5. 2 Mg( s ) + O 2 ( g ) → 2 MgO( s ) [121.5 g Mg & 64.0 g O 2 ] H 2 is the limiting reagent. CO 2 is the limiting reagent. O 2 is the limiting reagent. CH 4 is the limiting reagent. O 2 is the limiting reagent.

14 Limiting and Excess Reagents We can also use our knowledge of the limiting reagent to determine the quantity of a product in a reaction. Again, we use stoichiometry to do this.

15 Limiting and Excess Reagents For example, what is the maximum amount of CO 2 that will form from 64.0 g of methane, CH 4, and 224 g of oxygen gas? CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) M (g/mol) m (g) n (mol) ____ ____ ①② First, we calculate the number of mols of each reactant. ① n CH4 = m CH4 / M CH4 = (64.0 g)/(16.0 g/mol) = 4.00 mol 4.00 ② n O2 = m O2 / M O2 = (224 g)/(32.0 g/mol) = 7.00 mol 7.00

16 Limiting and Excess Reagents For example, what is the maximum amount of CO 2 that will form from 64.0 g of methane, CH 4, and 224 g of oxygen gas? CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) M (g/mol) m (g) n (mol) Next, we calculate the number of mols of O 2 needed to react with all of the CH n CH4 n O2 = 1 2 ⇒ 1 × n O2 = 2 × n CH4 ⇒ n O2 = 2 × 4.00 mol = 8.00 mol But, we only have 7.00 mols of O 2. We will run out of O 2 before we run out of CH 4. O 2 is the limiting reagent and it will control the amount of CO 2 produced.

17 Limiting and Excess Reagents For example, what is the maximum amount of CO 2 that will form from 64.0 g of methane, CH 4, and 224 g of oxygen gas? CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) M (g/mol) m (g) n (mol) Now, we use the amount of O 2 to calculate the maximum amount of CO 2 produced ⇒ 2 × n CO2 = 1 × n O2 ⇒ n CO2 = ½ × 7.00 mol = 3.50 mol ③ ④ n O2 n CO2 = 2 1 ③ = (3.50 mol) × (44.0 g/mol)= 154 g m CO2 = n CO2 × M CO2 ④

18 Limiting and Excess Reagents Find the limiting reagent and determine the maximum amount of product that will result. 1. N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g ) [42.0 g N 2 & 10.0 g H 2 ; find NH 3 ] 2. CaO( s ) + CO 2 ( g ) → CaCO 3 ( s ) [112 g CaO & 132 g CO 2 ; find CaCO 3 ] 3. 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O( l ) [12.0 g H 2 & 16.0 g O 2 ; find H 2 O] 4. CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O( l ) [64.0 g CH 4 & 288 g O 2 ; find CO 2 ] 5. 2 Mg( s ) + O 2 ( g ) → 2 MgO( s ) [121.5 g Mg & 96.0 g O 2 ; find MgO] N 2 is the limiting reagent; 51.0 g of NH 3 produced. CaO is the limiting reagent; 200. g of CaCO 3 produced. O 2 is the limiting reagent; 18.0 g of H 2 O produced. CH 4 is the limiting reagent; 176 g of CO 2 produced. Mg is the limiting reagent; 202 g of MgO produced.

19 Percent Yield When we do a stoichiometric calculation, we predict the maximum yield of products from the limiting reagents. For example, we calculate the maximum mass that we could expect from using the masses of reactants we used. This maximum yield is called the theoretical yield.

20 Percent Yield When we actually do the reaction in the laboratory, the amount of product we get is usually less than that theoretical yield. Why might that be so? We might have had trouble transferring the product to the balance. We might have lost some during heating or filtration or any other activity that involves handling the product.

21 Percent Yield We want to have a measure of the efficiency of the reaction carried out in the laboratory. We want something that tells us how close we came to the theoretical yield. We use percent yield. Percent yield is the ratio of the actual yield to the theoretical yield as a percentage. Percent yield = × 100% actual yield theoretical yield

22 Sample Problem 12.9 Calcium carbonate, which is found in sea shells, is decomposed by heating. What is the theoretical yield of CaO if 24.8 g of CaCO 3 is heated? The balanced chemical equation is: CaCO 3 ( s ) → CaO( s ) + CO 2 ( g ) ∆ M (g/mol) m (g) 24.8 n (mol) ① ② ① n CaCO3 = m CaCO3 M CaCO3 = 24.8 g g/mol = mol ②=②= n CaO n CaCO3 1 1 ⇒ n CaO = n CaCO3 = mol ③ m CaO = n CaO × M CaO ③ = (0.248 mol)(56.1 g/mol)= 13.9 g 13.9

23 Sample Problem What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? CaCO 3 ( s ) → CaO( s ) + CO 2 ( g ) From Sample Problem 12.9, the theoretical yield is 13.9 g. The actual yield is 13.1 g. ∆ Percent yield = × 100% theoretical yield actual yield Percent yield = × 100% 13.9 g 13.1 g = 94.2%

24 Summary In any chemical reaction, the amount of product is limited by the amount of reactants. When we have a reaction with more than one reactant, the reactant that runs out first is called the limiting reagent. The reactant that does not run out is call the excess reagent. We can also use our knowledge of the limiting reagent to determine the quantity of a product in a reaction.

25 Summary We want to have a measure of the efficiency of the reaction carried out in the laboratory. We want something that tells us how close we came to the theoretical yield. We use percent yield. Percent yield is the ratio of the actual yield to the theoretical yield as a percentage.


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