Stoichiometry Ch 10 10 1 Stoichiometry Mass amt

Stoichiometry Ch. 10

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

Mass to Mol Reminder Convert 3. 5 g of Na. OH to mols Na. OH 1. List known 3. 5 g Na. OH 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 Na. OH x 1 mol Na. OH = 0. 09 mol Na. OH 40 g Na. OH

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

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 + 3 H 2 2 NH 3

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

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

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

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

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 H 2 O = 14. 63 g H 2 O

Using Density • Density (D) = mass (m) / volume (V) • Units: g/m. L • Convert from g m. L or m. L g

Density Practice Calculate the mass of Li. OH used to obtain 1500 m. L of water. (Hint: DH 2 O = 1. 00 g/m. L) CO 2 + 2 Li. OH Li 2 CO 3 + H 2 O 1. Start w/ known & use density to convert into grams 1500 m. L H 2 O x 1. 00 g H 2 O 1 m. L H 2 O

Density Practice 2. Proceed w/ mass to mass calc. 1500 m. L H 2 O x 1. 00 g H 2 O x 1 mol H 2 O x 1 m. L H 2 O 18. 02 g H 2 O 2 mol Li. OH x 23. 95 g Li. OH = 3989 g Li. OH 1 mol H 2 O 1 mol Li. OH

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

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

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

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

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

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

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

% Yield Practice When 0. 835 mol Li. OH is reacted w/ excess KCl, the actual yield of Li. Cl is 16 g. What is the % yield? Li. OH + KCl Li. Cl + KOH

% Yield Practice 1. Calculate theor. yield 0. 835 mol Li. OH x 1 mol Li. Cl x 42. 44 g Li. Cl = 35. 4 g Li. Cl 1 mol Li. OH 1 mol Li. Cl 2. Calculate the % yield = act. X 100 = 16 g Li. Cl x 100 = 45 % theor. 35. 4 g Li. Cl