Presentation is loading. Please wait.

# Chapter 5 Chemical Reactions

## Presentation on theme: "Chapter 5 Chemical Reactions"— Presentation transcript:

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

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.

Chemical Reaction for CO2
Reactants Product O2 (g) CO2 (g) C(s)

+ The Limiting Reactant C(s) O2(g) CO2(g) + unreacted O2
Mixture of reactants + C(s) O2(g) CO2(g) + unreacted O2

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 O2? 2NO+O2=>2NO2 ; NO or O2

Solution To answer this question, calculate the grams of NO2 needed to react fully with 79.6 grams of NO and 59.5 grams of O2, by using the balanced equation. 79.6 g of NO 1 mol of NO 2 mol of NO2 46 grams of NO2 X X X 30.0 g of NO 2 mol of NO 1 mol of NO2 = grams of NO2 59.5 g of O2 1 mol of O2 2 mol of NO2 46 grams of NO2 X X X 32.0 g of O2 1 mol of O2 1 mol of NO2 = grams of NO2 There is less NO2 with NO than O2, therefore, the limiting reactant is NO

Learning Check For the balanced equation shown below, if 19.1 grams of CH5N were reacted with 88.1 grams of O2, how many grams of CO2 would be produced, using the limited reactant to determine the quantity of a product that should be produced ? 4CH5N + 11O2 => 4CO2 + 10H2O + 4NO

Solution To answer this question, calculate the grams of NO2 needed to react fully with 19.1 grams of CH5N and 88.1 grams of O2, by using the balanced equation. 19.1 g of CH5N 1 mol of CH5N 4 mol of CO2 44 g of CO2 X X X 31.0 g of CH5N 4 mol of CH5O 1 mol of NO2 = 21.1 grams of CO2 88.1 g of O2 1 mol of O2 4 mol of CO2 44 grams of CO2 X X X 32.0 g of O2 11 mol of O2 1 mol of CO2 = 44.1 grams of CO2 There is 21.1 g of CO2 produced with CH5N than O2, which is the limiting reactant

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 5.10. Actual yield % yield = X 100 Theoretical yield

Learning Check A chemist wants to produce urea (N2CH4O) by reacting ammonia (NH3) and carbon dioxide (CO2). The balanced equation for the reaction is 2NH3(g) + CO2 (g)  N2CH4O(s) + H2O(l) The chemist reacts 5.11 g NH3 with excess CO2 and isolates 3.12 g of solid N2CH4O. Calculate the percentage yield of the experiment.

Solution To answer this question, calculate first the theoretical yield of N2CH4O that should be made. Then use the actual yield to calculate the percentage yield. 5.11 g of NH3 1 mol of NH3 1 mol of N2CH4O X 60.1 g of N2CH4O X X 17.0 g of NH3 2 mol of NH3 1 mol of N2CH4O = 9.03 grams of N2CH4O is the theoretical yield The actual yield was 3.12 g of N2CH4O, so the % yield is Actual yield % yield = 3.12 g X 100 = X 100 = 34.6% Theoretical yield 9.03 g

Download ppt "Chapter 5 Chemical Reactions"

Similar presentations

Ads by Google