1. Stoichiometry Ch.10

2. (10-1) Stoichiometry Mass & amt relationships b/w reactants & products –Conversions b/w grams & moles Always begin w/ a balanced eq.!

3. Mass to Mol Reminder Convert 3.5 g of NaOH to mols NaOH 1.List known 3.5 g NaOH 2.Calculate molar mass 22.99 g/mol Na + 16 g/mol O + 1.01 g/mol H = 40 g/mol 3.Set up problem & solve 3.5 g NaOH x 1 mol NaOH = 0.09 mol NaOH 40 g NaOH

4. Mol to Mass Reminder How many grams are in 6.4 mols O 2 ? 1.List known 6.4 mols O 2 2.Calculate molar mass (2) (16 g/mol O) = 32 g/mol O 2 3.Set up problem & solve 6.4 mols O 2 x 32 g O 2 = 204.8 g O 2 1 mol O 2

5. Mol to Mol Reminder If 2.00 mol N 2 is reacting w/ a sufficient amt of H 2, how many mols of NH 3 will be produced? N 2 + H 2  NH 3 1.Balance the chemical eq. N 2 + 3H 2  2 NH 3

6. Mol to Mol Reminder 2.Find the mole ratio 1 mol N 2 : 2 mol NH 3 3.Set up problem (begin w/ known) & solve 2.00 mol N 2 x 2 mol NH 3 = 4 mol NH 3 1 mol N 2

7. Conversions g of A  mol of A  mol of B  g of B Mole ratio (coeff.from bal.eq.) Molar Mass (from PT) Molar Mass (from PT)

8. Mass to Mass Practice How many grams of H 2 O are produced when 13 g O 2 combine w/ sufficient H 2 ? H 2 + O 2  H 2 O 1.Balance the chemical eq. 2 H 2 + O 2  2 H 2 O

9. Mass to Mass Practice 2.Calculate molar mass of known & unknown O 2 = 32 g/mol H 2 O = 18.02 g/mol 3.Find mole ratio 1 mol O 2 : 2 mol H 2 O

10. Mass to Mass Practice 4. Set up problem (begin w/ known) & solve 13 g O 2 x 1 mol O 2 x 2 mol H 2 O x 18.02 g H 2 O 32 g O 2 1 mol O 2 1 mol H 2 O = 14.63 g H 2 O

11. Using Density Density (D) = mass (m) / volume (V) Units: g/mL Convert from g  mL or mL  g

12. Density Practice Calculate the mass of LiOH used to obtain 1500 mL of water. (Hint: D H 2 O = 1.00 g/mL) CO 2 + 2LiOH  Li 2 CO 3 + H 2 O 1.Start w/ known & use density to convert into grams 1500 mL H 2 O x 1.00 g H 2 O 1 mL H 2 O

13. Density Practice 2.Proceed w/ mass to mass calc. 1500 mL H 2 O x 1.00 g H 2 O x 1 mol H 2 O x 1 mL H 2 O 18.02 g H 2 O 2 mol LiOH x 23.95 g LiOH = 3989 g LiOH 1 mol H 2 O 1 mol LiOH

14. (10-2) Excess Reactant Extra left over after rxn Limiting reactant: completely consumed –Limits amt of other reactants used –Determines max amt of product

16. Limiting Reactant Practice CO combines w/ H 2 to produce CH 3 OH. If you had 152.5 g CO & 24.5 g H 2 what mass of CH 3 OH could be produced? 1.Write a bal. eq. CO + 2 H 2  CH 3 OH

17. Limiting Reactant Practice 2.Convert reactants to mols present 152.5 g CO x 1 mol CO = 5.444 mol CO present 28.01 g CO 24.50 g H 2 x 1 mol H 2 = 12.1 mol H 2 present 2.02 g H 2

18. Limiting Reactant Practice 3.Using the reactants mol ratio, find how many mols needed 12.1 mol H 2 x 1 mol CO = 6.06 mol CO needed 2 mol H 2 CO present is not enough to react w/ all the H 2, so CO is limiting

19. Limiting Reactant Practice 4. Use limiting reactant to set up stoich. 5.444 mol CO x 1 mol CH 3 OH x 32.05 g CH 3 OH 1 mol CO 1 mol CH 3 OH = 174.5 g CH 3 OH

20. Theoretical Yield Calculated max amt of product possible –Stoich. w/ limiting reactant –What should happen Actual yield: measured amt of product experimentally produced –What does happen

21. Percentage Yield How efficient a rxn is –How close actual is to theoretical –Should be 100% % yield = actual x 100 theoretical

22. % Yield Practice When 0.835 mol LiOH is reacted w/ excess KCl, the actual yield of LiCl is 16 g. What is the % yield? LiOH + KCl  LiCl + KOH

23. % Yield Practice 1. Calculate the theor. yield 0.835 mol LiOH x 1 mol LiCl x 42.44 g LiCl = 35.4 g LiCl 1 mol LiOH 1 mol LiCl 2.Calculate the % yield % yield = act. X 100 = 16 g LiCl x 100 = 45 % theor. 35.4 g LiCl