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1 Chapter 5 Chemical Reactions 5.10 & 5.11 The Limiting Reactant and Reaction Yield.

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Presentation on theme: "1 Chapter 5 Chemical Reactions 5.10 & 5.11 The Limiting Reactant and Reaction Yield."— Presentation transcript:

1 1 Chapter 5 Chemical Reactions 5.10 & 5.11 The Limiting Reactant and Reaction Yield

2 2 Definitions Limiting-reactant principle – The maximum amount of product possible from a reaction is determined by the amount of reactant present in the least amount, based on its reaction coefficient and molecular weight. Limiting reactant – the reactant present in a reaction in the least amount, based on its reaction coefficients and molecular weight. It is the reactant that determines the maximum amount of product that can be formed.

3 3 Chemical Reaction for CO 2 Reactants Product C(s) O 2 (g) CO 2 (g)

4 4 + C(s) O 2 (g) Mixture of reactants CO 2 (g) + unreacted O 2 The Limiting Reactant

5 5 Learning Check For the balanced equation shown below, what would be the limiting reagent if 79.6 grams of NO were reacted with 59.5 grams of O 2 ? 2NO+O 2 =>2NO 2 ; NO or O 2

6 6 Solution To answer this question, calculate the grams of NO 2 needed to react fully with 79.6 grams of NO and 59.5 grams of O 2, by using the balanced equation g of NO X 1 mol of NO 30.0 g of NO X 2 mol of NO 2 2 mol of NO X 46 grams of NO 2 1 mol of NO 2 = grams of NO g of O 2 X 1 mol of O g of O 2 X 2 mol of NO 2 1 mol of O 2 X 46 grams of NO 2 1 mol of NO 2 = grams of NO 2 There is less NO 2 with NO than O 2, therefore, the limiting reactant is NO

7 7 Learning Check For the balanced equation shown below, if 19.1 grams of CH 5 N were reacted with 88.1 grams of O 2, how many grams of CO 2 would be produced, using the limited reactant to determine the quantity of a product that should be produced ? 4CH 5 N + 11O 2 => 4CO H 2 O + 4NO

8 8 Solution To answer this question, calculate the grams of NO 2 needed to react fully with 19.1 grams of CH 5 N and 88.1 grams of O 2, by using the balanced equation g of CH 5 N X 1 mol of CH 5 N 31.0 g of CH 5 N X 4 mol of CO 2 4 mol of CH 5 O X 44 g of CO 2 1 mol of NO 2 = 21.1 grams of CO g of O 2 X 1 mol of O g of O 2 X 4 mol of CO 2 11 mol of O 2 X 44 grams of CO 2 1 mol of CO 2 = 44.1 grams of CO 2 There is 21.1 g of CO 2 produced with CH 5 N than O 2, which is the limiting reactant

9 9 Percent Yield Percentage yield – the percentage of the theoretical amount of a product actually produced by a reaction. Actual yield – the mass product obtained in an experiment. Theoretical yield – the mass calculated according to the methods used in section 5.9 and % yield = Actual yield Theoretical yield X 100

10 10 Learning Check A chemist wants to produce urea (N 2 CH 4 O) by reacting ammonia (NH 3 ) and carbon dioxide (CO 2 ). The balanced equation for the reaction is 2NH 3 (g) + CO 2 (g)  N 2 CH 4 O(s) + H 2 O(l) The chemist reacts 5.11 g NH 3 with excess CO 2 and isolates 3.12 g of solid N 2 CH 4 O. Calculate the percentage yield of the experiment.

11 11 1 mol of N 2 CH 4 O Solution To answer this question, calculate first the theoretical yield of N 2 CH 4 O that should be made. Then use the actual yield to calculate the percentage yield g of NH 3 X 1 mol of NH g of NH 3 X 1 mol of N 2 CH 4 O 2 mol of NH 3 X = 9.03 grams of N 2 CH 4 O is the theoretical yield 60.1 g of N 2 CH 4 O % yield = Actual yield Theoretical yield X100 = The actual yield was 3.12 g of N 2 CH 4 O, so the % yield is 3.12 g 9.03 g X100 = 34.6%


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