38.1 Limiting and Excess Reagents What would happen in the following situation?figure 8.1, page 2962 sandwiches6 sandwiches5 sandwichesOnly 2 sandwiches be made because …..
48.1 Limiting and Excess Reagents toast is the limiting reagentDo Thought Lab 8.1, page 296
58.1 Limiting and Excess Reagents limiting reagent is completely consumed in in a particular chemical reactionexcess reagent is partially consumed in a particular chemical reactioneven the identity of products of a chemical reaction are sometimes determined by whether a given reactant is limiting or excess
68.1 Limiting and Excess Reagents how to identify limiting reagent:easiest way – find moles of each reactant, use to find which produces the least number of moles of product –any product!
78.1 Limiting and Excess Reagents Example: Practice Problem 6, page 299C3H6(g) + 2 NH3(g) + 2 O2(g) C3H3N(g) + HCN(g) + 4 H2O(g)n1 1.0 kgn2 600 gn3pick a product – it doesn’t matter which, and find out which makes least number of moles of product I’ll use C3H3N and call it n3limiting reagent is not necessarily the one with smaller mass
88.1 Limiting and Excess Reagents Once you’ve identified the limiting reagent you can do stoichiometry to calculate expected yieldsExample: Practice Problem 7&8, page 3037. identify the limiting reagent – find which makes least moles of Mg3(PO4)(s)limiting3 Mg(NO3)2(aq) + 2 Na3PO4(aq) Mg3(PO4)2(s) + 6 NaNO3(aq)n mL 0.5 mol/Ln mL 1.2 mol/Ln3
98.1 Limiting and Excess Reagents 8. Calculate the mass of Mg3(PO4)2(s) formedn = mol x g/mol = 4 gWhat would happen if you used the wrong substance as limiting reagent?You would calculate a larger mass of Mg3(PO4)2(s)
108.1 Limiting and Excess Reagents Worksheet BLM 8.1.3Worksheet BLM 8.1.5, questions 1-3 only
118.2 Predicted and Experimental Yields Predicted or theoretical yield – determined by stoichiometryExperimental or actual yield – what you end up gettingLab 8A, page 300
138.2 Predicted and Experimental Yields Example: question 4 page 311
148.2 Predicted and Experimental Yields 2 NaCl(aq) + 1 Pb(NO3)2(aq) PbCl2(s) + 2 NaNO3(aq)n gn gprecipitate n3 m=?limitingWorksheet BLM 8.2.1
158.2 Predicted and Experimental Yields b)Worksheet BLM 8.2.1
168.3 Acid-Base Titration Titration Set-up: Titration talk: fig 8.5, page 312Titration talk:“titration of withsampletitrant”Point where erlenmeyer flask contains stoichiometrically equivalent moles of acid and base:equivalence pointif indicator is properly chosen, endpoint occurs at equivalence pointPoint where indicator changes colour:endpoint
178.3 Acid-Base Titrationstandardizing: doing a titration to find the concentration of a titrant solution to be used in further analysesHCl(aq) needs to be standardized since pure HCl is a gas and escapes from solutionNaOH(aq) needs to be standardized since its solutions absorb CO2(g) from the air causing its pH to droppopular titrants
188.3 Acid-Base Titration endpoints observed using acid-base indicators indicators are weak acid/base pairs where the 2 members have different colourschart page 10 of Data Booklet shows indicator acid/base pairsHIn(aq) H+(aq) + In‾(aq)colour 1colour 2
208.3 Acid-Base Titration Indicators used to show endpoint Discuss questions 4-6, page 314
218.3 Acid-Base TitrationTitration calculations – solution stoichiometryExample: Practice Problem 21, page 315Questions states that “a student titrates HCl(aq) with NaOH(aq)” Which is the titrant?NaOH(aq)HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)n2 v=20.00 mL c=?n1 v=( ) mL c=0.150 mol/Lto be continued …….
228.3 Acid-Base Titration Practice Problem 21, page 315, continued Practice Problem 22, page 315 states that “a student uses NaOH(aq) to titrate HNO3(aq)”Which is the titrant?NaOH(aq)Note that the base isn’t always the titrantWorksheet BLM 8.3.3, omit 1a, b