lect_8oct20f09

lect_8oct20f09 - 1 Outline 8 20-OCT-2009 1.Review Questions...

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Unformatted text preview: 1 Outline 8 20-OCT-2009 1.Review- Questions? Stoichiometry needs balanced eqn & molar masses Limiting reagent (Least amount, ratio, ICE) 2. % yield 3. Solution Concentrations, mol/L Another way to get moles! # mol = M(mol/L)*V(L) 4. Solution Stoichiometry 2 Limiting Reagent : Ex. 1 (s4,cont) Ex: ICE Approach: let x = # mol rxn Zn + 2 HCl → ZnCl 2 + H 2 I nitial 0.30 0.52 C hange-x-2x x x E nd (0.30-x) (0.52-2x) x x Explanation: I (nitial): Amounts of each at beginning, can be grams or mol; if grams convert to mol. C (hange): mol of each used up (-) or made. # mol Zn reacted = x (mol rxn)* (1 mol Zn/mol rxn) # mol HCl reacted = x (mol rxn)* (2 mol HCl/mol rxn) # mol ZnCl 2 made = x (mol rxn)* (1 mol ZnCl 2 /mol rxn) # mol H 2 made = x (mol rxn)* (1 mol H 2 /mol rxn) 3 Limiting Reagent : Ex. 1 (s5 cont) Ex: ICE Approach: Let x = # mol rxn Zn + 2 HCl → ZnCl 2 + H 2 I nitial 0.30 0.52 C hange-x-2x x x E nd (0.30-x) (0.52-2x) x x Explanation: (continued) E (nd): these are mol of species at end of rxn; (initial – change) # mol Zn remaining = (0.30-x) # mol HCl remaining = (0.52-2x) # mol ZnCl 2 made = x # mol H 2 made = x End explanation. Now to finish the problem, next slide. 4 Limiting Reagent : Ex. 1 (s6 cont) Ex: ICE Approach: Let x = # mol rxn Zn + 2 HCl → ZnCl 2 + H 2 I nitial 0.30 0.52 C hange-x-2x x x E nd (0.30-x) (0.52-2x) x x If Zn limiting (used up first) then (0.30-x) =0 so x = 0.30 If HCl limiting (used up first) then (0.52-2x) =0 so x = 0.26 Smallest # mol rxn (x) is from HCl (0.26 vs. 0.30), so mol HCl remaining = , used up (limiting) mole H 2 made = x mol rxn* (1mol H 2 /mol rxn) = 0.26 mol H 2 mole Zn remaining = 0.30 - x = 0.30 - 0.26 = 0.04 mol Zn mole ZnCl 2 made = x = 0.26 mol ZnCl 2 5 Limiting Reagent : Ex. 2(s1) Given: A rxn vessel contains 0.25 mol KO 2 and 0.15 mol water. 4 KO 2 (s) + 2 H 2 O → 4 KOH (s) + 3 O 2 (g) Find: (a) limiting reagent (b) mol oxygen made (c) mol of other species at end. Soln: get (a) mol O 2 (or KOH) that could be made by each or (b) ratios or (c) use ICE 6 Limiting Reagent : Ex. 2 (s2) Given: A rxn vessel contains 0.25 mol KO 2 and 0.15 mol water. 4 KO 2 (s) + 2 H 2 O → 4 KOH (s) + 3 O 2 (g) Soln: (a) Least amount . mol O 2 from each reactant # mol O 2 from KO 2 = 0.25 mol KO 2 *(3 mol O 2 /4 mol KO 2 )= 0.19 mol O 2 # mol O 2 from H 2 O = 0.15 mol H 2 O *(3 mol O 2 /2 mol H 2 O)= 0.23 mol O 2 So KO 2 is limiting (gives smallest amount) Or use Ratio (next slide) 7 Limiting Reagent : Ex. 2 (s3) Given: A rxn vessel contains 0.25 mol KO 2 and 0.15 mol water. 4 KO 2 (s) + 2 H 2 O → 4 KOH (s) + 3 O 2 (g) Soln: (b) Ratio. Have: 0.25 mol KO 2 /0.15 mol H 2 O = 1.7 mol KO 2 / mol H 2 O Chem Eqn ratio requires: 4 mol KO 2 /2 mol H 2 O = 2 mol KO 2 / mol H 2 O So limiting reagent is KO 2 (not enough to react completely with H 2 O) 8 Limiting Reagent : Ex. 2 (s4) Given: A rxn vessel contains 0.25 mol KO 2 and 0.15 mol water. 4 KO 2 (s) + 2 H 2 O → 4 KOH (s) + 3 O 2 (g) Soln: Rest of Calculations are same:...
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lect_8oct20f09 - 1 Outline 8 20-OCT-2009 1.Review Questions...

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