Presentation on theme: "Chapter 5 Chemical Reactions"— Presentation transcript:
1 Chapter 5 Chemical Reactions 5.10 & 5.11The Limiting Reactant and Reaction Yield
2 DefinitionsLimiting-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 Chemical Reaction for CO2 Reactants ProductO2 (g)CO2 (g)C(s)
4 + The Limiting Reactant C(s) O2(g) CO2(g) + unreacted O2 Mixture of reactants+C(s)O2(g)CO2(g) + unreacted O2
5 Learning CheckFor 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
6 SolutionTo 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 NO1 mol of NO2 mol of NO246 grams of NO2XXX30.0 g of NO2 mol of NO1 mol of NO2= grams of NO259.5 g of O21 mol of O22 mol of NO246 grams of NO2XXX32.0 g of O21 mol of O21 mol of NO2= grams of NO2There is less NO2 with NO than O2, therefore, the limiting reactant is NO
7 Learning CheckFor 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
8 SolutionTo 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 CH5N1 mol of CH5N4 mol of CO244 g of CO2XXX31.0 g of CH5N4 mol of CH5O1 mol of NO2= 21.1 grams of CO288.1 g of O21 mol of O24 mol of CO244 grams of CO2XXX32.0 g of O211 mol of O21 mol of CO2= 44.1 grams of CO2There is 21.1 g of CO2 produced with CH5N than O2, which is the limiting reactant
9 Percent YieldPercentage 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 =X100Theoretical yield
10 Learning CheckA chemist wants to produce urea (N2CH4O) by reacting ammonia (NH3) and carbon dioxide (CO2). The balanced equation for the reaction is2NH3(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.
11 SolutionTo 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 ofNH31 mol of NH31 mol of N2CH4OX60.1 g of N2CH4OXX17.0 g of NH32 mol of NH31 mol of N2CH4O= 9.03 grams of N2CH4O is the theoretical yieldThe actual yield was 3.12 g of N2CH4O, so the % yield isActual yield% yield =3.12 gX100 =X100 =34.6%Theoretical yield9.03 g