1. SOL Review 6 Stoichiometry

2. Consider: 4NH 3 + 5O 2  6H 2 O + 4NO Many conversion factors exist: 4 NH 3 6 H 2 04NO 5O 2 (and others) 5 O 2 4 NO4 NH 3 6 H 2 0 The Coefficients of a balanced chemical reaction can be used to write conversion factors called MOLE RATIOS. With these ratios, you can relate moles of reactants to moles of product.

3. Consider: 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 0.176 mol of O 2 are used? 0.211 mol H 2 O 6 mol H 2 O 5 mol O 2 x # mol H 2 O=0.176 mol O 2 = Notice that a correctly balanced equation is essential to get the right answer!

4. Consider: 4NH 3 + 5O 2  6H 2 O + 4NO How many grams of NO are produced if we start with 2.0 moles of O 2 ? We find out how many moles of NO, then convert moles of NO to grams of NO. 2 moles O 2 x 4 moles NO x 30.0 g = 48 g NO 5 moles O 2 1 mole NO

5. Conversions To Different Units Of Measurement We can also go from liters of a reactant to moles of a product 4NH 3 + 5O 2  6H 2 O + 4NO 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of NO are produced if we start with 3.00 liters of O 2 ? 3 L O 2 x 1 mole x 4 mol NO =.107 mole NO 22.4L 5 mol O 2 22.4L 5 mol O 2

6. Limiting Reagent: The reactant that limits the amount of the product that can be formed in the reaction. This is the reactant that is completely used up in a reaction. Excess Reagent – reactant that is not completely used up.

7. Percent yield Percent yield is method for comparing the amount of product made in the lab to the expected result. Reasons why the experimental yield is not the same as the theoretical yield: Not all reactions go to completion. Some may have a “side” reaction and not produce the desired result. Sometimes it is hard to purify the product.

8. Percent yield = __Actual yield x 100% Theoretical yield The actual yield is the amount made in the lab. The theoretical yield is how much should have been made according to a mass-mass problem using a balanced equation (stoichiometry calculation).

9. CaCO 3  CaO + CO 2 1. What is the theoretical yield of CaO if 24.8g CaCO 3 is heated? MM CaO = 56 g/mol, MM CaCO 3 = 100 g/mol 24.8 g CaCO 3 x 1 mol CaCO 3 x _1 mol CaO x ___56 g = 100 g 1 mol CaCO 3 1 mol CaO 13.9 g CaO 2. What is the percent yield if 13.1 g is actually produced? 13.1 g x 100% = 94.2% yield 13.9 g