lect_8oct20f09

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

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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:...
View Full Document

{[ snackBarMessage ]}

### Page1 / 38

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

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online