1. Theoretical Yield The amount of product(s) calculated using stoichiometry problems

2. Actual Yield The amount of product actually recovered when the reaction is done in the lab

3. Percent Yield (Actual yield/Theoretical Yield) x 100

4. A reaction between solid sulfur and oxygen produces sulfur dioxide. 1S 6 (s) + 6O 2 (g)  6SO 2 (g)

5. The reaction started with 384 grams of S 6 (s). Assume an unlimited supply of oxygen. What is the theoretical (predicted) yield and the percent yield if only 680 grams of sulfur dioxide are produced? Find moles of S 6 384 g S 6 ( 1 moles S 6 / 192 g S 6 ) = 2 moles S 6 Find moles of SO 2 2 moles S 6 (6 moles SO 2 /1 mole S 6 ) = 12 moles SO 2

6. Find theoretical grams SO 2 12 moles SO2 (64 g SO2/1mole) = 768 g SO2 Find Percent Yield (680 g / 768g) x100 = 89%

7. Zinc reacts with copper sulfate in a single replacement reaction as follows Zn (s) + CuSO 4 (aq)  ZnSO 4 (aq) + Cu (s) 50.00 grams of zinc metal were added to excess copper sulfate dissolved in a water solution. 42.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment

8. Step 1: Find the theoretical mass of Cu Mass of Cu = – 50.00g Zn(mol/65.38gZn) = 0.7648 mol Zn – 0.7648 mol Zn (1 molCu/1 mol Zn) = 0.7648 mol Cu – 0.7648 mol Cu (63.55g Cu/mol) = 48.60 g Cu

9. Step 2: Find Percent Yield % Yield = (42.50 g Cu/48.60 g Cu)x 100 = 87.44%

10. Silver can produced by reacting silver nitrate with magnesium in the following reaction Mg(s) + 2 AgNO 3 (aq)  Mg(NO 3 ) 2 (aq) + 2 Ag (s)

11. How much Silver can be recovered by reacting a silver nitrate solution with 50.00 grams of powdered magnesium. 50.00g Mg(mol/24.31 g Mg) = 2.057 mol Mg 2.057 mol Mg ( 2mol Ag/ 1 mol Mg) = 4.114 mol Ag 4.114 mol Ag (107.9 g/mol) =443.8 g Ag

12. Assume that 95% of the silver can be recovered. 443.8 g theoretically produced. 443.8 g (0.95) = 421.58 g Ag actually produced

13. MOLARITY Concentration

14. Solutions Solute: the substance being dissolved Solvent: the substance doing the dissolving

15. Units of Concentration Molarity – Ratio of – “Amount” of Solute/ “Amount” of Solution

16. Molarity Moles of solute/L of solution M

17. Prepare 500mL of a 1M sucrose solution

18. Prepare 250 mL of a 6M solution of dye 4 drops = 1 mole