1. GOOD AFTERNOON! Prepare to take notes. You will be allowed to use these notes on the test next week! You will need: 1.Something to write on 2.Something to write with 3.A periodic table 4.A calculator (you may have to share with your neighbor)

2. STOICHIOMETRY Calculations using chemical reactions

3. MOLE-TO-MOLE RATIO The coefficients of a balanced chemical reaction can be used to show the relationship between the products and or reactants of a chemical reaction. This is true because the units of a balanced chemical reaction are moles. The mole-to-mole ratio is a ratio between reactants and products which can be used to convert between any two compounds in a chemical reaction. The mole-to-mole ratio is used in any problem that converts between amounts of two different substances (compounds or elements).

4. MOLE-TO-MOLE CONVERSION PROBLEM EXAMPLE: According to the following balanced chemical equation, how many moles of Na 2 SO 4 can be produced from the reaction of 14.3 moles of NaOH with excess H 2 SO 4 ? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O 14.3 moles NaOH  ? moles Na 2 SO 4 = 14.3 moles Na 2 SO 4 = 7.15 moles Na 2 SO 4 2 14.3 moles NaOH1 mole Na 2 SO 4 2 moles NaOH

5. GRAM-TO GRAM STOICHIOMETRY: Because the units of chemical reaction coefficients are moles and not grams, you must always convert to moles before converting from units of one compound to units of another compound

6. GRAM-TO-GRAM STOICHIOMETRY EXAMPLE PROBLEM: Example: How many grams of NaOH are required to react completely with 28.2 grams of H 2 SO 4 according to the following balanced reaction? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 23.0 g NaOH 28.2 g H 2 SO 4 1 mol H 2 SO 4 2 mol NaOH40.0 g NaOH 98.1 g H 2 SO 4 1 mol H 2 SO 4 1 mol NaOH

7. GRAM-TO-MOLE STOICHIOMETRY EXAMPLE PROBLEM: Example: How many moles of NaOH are required to react completely with 28.2 grams of H 2 SO 4 according to the following balanced reaction? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 0.575 moles NaOH 28.2 g H 2 SO 4 1 mol H 2 SO 4 2 mol NaOH 98.1 g H 2 SO 4 1 mol H 2 SO 4

8. MOLE-TO-GRAM STOICHIOMETRY EXAMPLE PROBLEM: Example: How many grams of NaOH are required to react completely with 14.3 moles of H 2 SO 4 according to the following balanced reaction? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 1,144 g NaOH 14.3 mol H 2 SO 4 2 mol NaOH40.0 g NaOH 1 mol H 2 SO 4 1 mol NaOH

9. LIMITING REACTANT, EXCESS REACTANT, AND THEORETICAL YIELD: The limiting reactant is the reactant that runs out first. It limits the reaction more than any other reactant. Excess reactants are the reactants that are not used up in a chemical reaction. They are left over. By definition, they are all reactants other than the limiting reactant. Theoretical yield is the amount of product you should be able to make using the given amount of each reactant. It is limited by the limiting reactant and is determined through calculations using the limiting reactant.

10. MOLE-TO-MOLE LIMITING REACTANT PROBLEM: According to the following balanced chemical equation, how many moles of Na 2 SO 4 can be produced from the reaction of 14.3 moles of NaOH with 12.4 moles of H 2 SO 4 ? What is the limiting reactant, excess reactant, and theoretical yield? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 7.15 moles Na 2 SO 4 = 12.4 moles Na 2 SO 4 14.3 moles NaOH1 mole Na 2 SO 4 2 moles NaOH 12.4 moles H 2 SO 4 1 mole Na 2 SO 4 1 mole H 2 SO 4

11. GRAM-TO-GRAM LIMITING REACTANT PROBLEM: According to the following balanced chemical equation, how many grams of Na 2 SO 4 can be produced from the reaction of 24.3 grams of NaOH with 22.4 grams of H 2 SO 4 ? What is the limiting reactant, excess reactant, and theoretical yield? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 43.2 g Na 2 SO 4 = 32.4 g Na 2 SO 4 24.3 g NaOH1 mol NaOH1 mol Na 2 SO 4 142.1 g Na 2 SO 4 40.0 g NaOH2 mol NaOH1 mol Na 2 SO 4 22.4 g H 2 SO 4 1 mol H 2 SO 4 1 mol Na 2 SO 4 142.1 g Na 2 SO 4 98.1 g H 2 SO 4 1 mol H 2 SO 4 1 mol Na 2 SO 4

12. HOW MUCH EXCESS IS LEFT OVER? 1.Determine which reagent is the limiting reagent 2.Find out how much of the excess reagent was used in the reaction 3.Subtract the amount used (from step 2) from the amount available (given in the problem).

13. GRAM-TO-GRAM EXCESS REACTANT PROBLEM: According to the following balanced chemical equation, how many grams of Na 2 SO 4 can be produced from the reaction of 24.3 grams of NaOH with 22.4 grams of H 2 SO 4 ? What is the limiting reactant, excess reactant, and theoretical yield? How much excess reagent is left over? Step 1: H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O = 43.2 g Na 2 SO 4 = 32.4 g Na 2 SO 4 24.3 g NaOH1 mol NaOH1 mol Na 2 SO 4 142.1 g Na 2 SO 4 40.0 g NaOH2 mol NaOH1 mol Na 2 SO 4 22.4 g H 2 SO 4 1 mol H 2 SO 4 1 mol Na 2 SO 4 142.1 g Na 2 SO 4 98.1 g H 2 SO 4 1 mol H 2 SO 4 1 mol Na 2 SO 4

14. GRAM-TO-GRAM EXCESS REACTANT PROBLEM: According to the following balanced chemical equation, how many grams of Na 2 SO 4 can be produced from the reaction of 24.3 grams of NaOH with 22.4 grams of H 2 SO 4 ? What is the limiting reactant, excess reactant, and theoretical yield? How much excess reagent is left over? H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O Step 2: = 18.3 g NaOH used Step 3: 24.3 g NaOH (available) -18.3 g NaOH (used) = 6.00 g NaOH left 22.4 g H 2 SO 4 1 mol H 2 SO 4 2 mol NaOH40.0 g NaOH 98.1 g H 2 SO 4 1 mol H 2 SO 4 1 mol NaOH

15. PERCENT YIELD Percent yield is a way to know how efficient a chemical reaction is by comparing the actual results of a reaction to the predicted yield (also known as theoretical yield) of a reaction. The formula for percent yield is % yield = x100 Actual yield will be given in the problem on a test or quiz Actual Yield Theoretical Yield

16. PERCENT YIELD EXAMPLE PROBLEM For the balanced equation shown below, if the reaction of 20.7 grams of CaCO 3 produces 6.81 grams of CaO, what is the percent yield? CaCO 3  CaO+CO 2 = 11.6 g CaO % yield = x 100 = 58.7% 20.7 g CaCO 3 1 mol CaCO 3 1 mol CaO56.1 g CaO 100.1 g CaCO 3 1 mol CaCO 3 1 mol CaO 6.81 g CaO (actual) 11.6 g CaO (theoretical)