Topic 83 StoichiometryLimiting and Excess Reagent 话题83的化学计量比的限制与过量的试剂

1、topic 8.3 stoichiometry: limiting and excess reagent calculations by kirsten page 320-327what is stoichiometry?lets take a closer look.la method of predicting or analyzing the quantities of the reactants and products participating in a chemical process lthree forms including gas stoichiometry, solut

2、ion stoichiometry, and gravimetric stoichiometry.example of stoichiometrylin a precipitation reaction, koh(aq) reacts with excess sn(no3)2(aq) to produce a precipitate. if the mass of precipitate is 2.57 g, what mass of koh(s) was present in the original solution?first step:balance equation:2koh(aq)

3、 + sn(no3) 2(aq) sn(oh)2(s) + 2kno3(aq) molesn2n1mass?2.57gmolar mass56.11 g/mol152.71 g/molsecond step:n1= 2.57 g x (1 mol/ 152.71 g) = 0.0168 moln2= 0.0168 mol x (2/1) = 0.0337 molm= 0.0337 mol x (56.11 mol/g) = 1.89gunderstanding chemical principleslconservation of mass in a chemical reactionlin

4、chemical reactions the mass is conserved, meaning that the mass of the products equals the mass of the reactants.lmass products = mass reactantslthere will always be an equal number of moles of each element before and after a reaction takes placeidentifying reagentsllimiting reagentslcompletely cons

5、umed in a chemical reactionlexcess reagentslmore is present than is necessary to react with the limiting reagentlthe limited reagent is the reagent that is being analyzed in a quantitative analysis where limiting and excess reagents are present.how to find limiting regent:lensure that all masses giv

6、en are in moles.ltake the molar amounts from each substance and divide by the coefficient of that substance in the balanced equation.lthe smaller number will be the limiting reagent. example 1:l300 ml of 0.100 mol/l of bacl2(aq) and 200 ml of 0.110 mol/l of na2co3(aq) are mixed. what is the limiting

7、 reagent in the reaction?answer: na2co3why?bacl2(aq) + na2co3(aq) baco3(s) + 2nacl(aq)with a 1:1 mole ratio of reactants he species present in least amount is the limiting reagent.nbacl2= 300ml x (0.100mol/l) =30.0 mmolnna2co3= 200 ml x (0.110mol/l)= 22.0 mmoltherefore na2co3(aq) is the limiting rea

8、gentexample 2:l100.0 g of iron (iii) chloride and 50.00g of hydrogen sulfide react. what is the limiting reagent?2fecl3 + 3h2s fe2s3 + 6hclanswer:liron(iii) chloride100g/ 162.204 g/mol=0.6165 mollhydrogen sulfide50.00g/ 34.081 g/mol=1.467 mol1.467 mol/ 3=0.489liron (iii) chloride is the limiting rea

9、gentexample 3:lin an experiment, 26.8 g of iron(iii) chloride in solution is combined with 21.5 g of sodium hydroxide in solution. which reactant is in excess, and by how much? what mass of each product will be obtained?answer:fecl3(aq) + 3naoh(aq) fe(oh)3(s) + 3nacl(aq) 26.8 g 21.5g m m162.20g/mol

10、40.00 g/mol 106.88 g/mol 58.44 g/molnfecl3 = 26.8 g x (1mol/ 162.20g0 =0.165 molnnaoh = 21.5 g x (1 mol / 40.00g)= 0.538 mol answer:nnaoh=0.165 mol x (3/1) =0.496nnaoh= 0.538 mol - 0.496 mol = 0.042 molmnaoh = 0.042 mol x (40.00g/1 mol) = 1.7ganswer:mfe(oh)3= 0.165 molfecl3 x (1 mol fe(oh)3/1 molfec

11、l3) x (106.88 gfe(oh)3/ 1 mol fe(oh)3) = 17.7 g fe(oh)3m nacl = 0.165 molfecl3 x (3 mol nacl/ 1 mol fecl3) x (58.44 g nacl / 1 mol nacl) = 29.0 nacltherefore sodium hydroxide is in excess by 1.7 g, the mass of iron (iii) hydroxide produced is 17.7 g and the mass of sodium chloride produced is 29.0g.

12、theoretical yields vs. actual yieldslthe theoretical yield of a chemical reaction is the amount of product formed if all of the limiting reagent reacts.lcalculated using stoichiometry.lthe actual yield is the actual quantity of products formed after a chemical reaction.lusually the theoretical yield

13、 will be greater than the actual yield that is produced. reasons for discrepancylthe actual yield of a chemical reaction is usually less than the theoretical yield for these reasonslpurity of chemicals being usedlerrors in measurementslexperimental factors that may have lead to loss of reactants% er

14、ror calculations:la reasonable quantity of reasonable excess reagent is 10%example 4:lyou decide to test the method of stoichiometry using the reaction of 2.00 g of copper(ii) sulfate in solution with an excess of sodium hydroxide in solution. what would be a reasonable mass of sodium hydroxide to use?cuso4(aq) + 2naoh(aq) cu(oh)2(aq) + na2so4(aq)2.00 g m159.62 g/mol 40.00 g/molanswer:ncuso4 = 2.00g x (1 mol/159.62 g) =0.0125 molnnaoh =0.0125 mol x (2/1) =0.0251 molmnaoh = 0.02