Presentation on theme: "#29 When 84.8 g of iron (III) oxide reacts with excess of carbon monoxide, iron is produced. Fe 2 O 3(s) + 3CO (g) 2Fe (s) + 3CO 2(g) What is the theoretical."— Presentation transcript:
#29 When 84.8 g of iron (III) oxide reacts with excess of carbon monoxide, iron is produced. Fe 2 O 3(s) + 3CO (g) 2Fe (s) + 3CO 2(g) What is the theoretical yield (how much iron do you expect to make?) of this reaction? 84.8 g Fe 2 O mole Fe 2 O g Fe 2 O 3 X = 59.3 g Fe XX 2 mole Fe 1mole Fe 2 O g Fe 1mole Fe
#30 When 5.00 g copper reacts with excess silver nitrate, silver metal and copper (II) nitrate are produced. What is the theoretical yield of silver in this reaction? Cu (s) + 2AgNO 3(aq) 2Ag (s) + Cu(NO 3 ) 2(aq) 5.00 g Cu 1 1 mole Cu g Cu X = 17.0 g Ag XX 2 mole Ag 1 mole Cu g Ag 1 mole Ag
#31 If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO 2(s) + 3C (s) SiC (s) + 2CO (g) What is the % yield in this reaction? 50.0 g SiO mole SiO g SiO 2 X = g SiC theoretical XX 1 mole SiC 1 mole SiO g SiC 1 mole SiC First calculate the theoretical yield: 27.9 g SiC X 100 = 83.6 % yield g SiC actual yield X 100 theoretical yield: Formula for % yield:
#32 If 15.0 g of nitrogen reacts with 15.0 g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield of this reaction? N 2 + 3H 2 2NH 3 calculate the theoretical yield for both nitrogen and hydrogen separately The one that produces the least ammonia determines the theoretical yield, then use that value to determine the % yield: 15.0 g N 2 1 X 1 mole N 2 28 g N 2 X 2 mole NH 3 1 mole N 2 X 17 g NH 3 1 mole NH 3 = 18.2 g NH g H 2 1 X XX 1 mole H g H 2 2 mole NH 3 3 mole H 2 17 g NH 3 1 mole NH 3 =85 g NH g NH g NH 3 X 100 = 57.7%
#33 In a chemical reaction, how does an insufficient quantity of a reactant affect the amount of product formed? If a reactant runs out, the reaction must stop. It is like building bicycles, if you run out of one piece you stop building bicycles.
#34 How can you gauge the efficiency of a reaction carried out in the laboratory? You can compare your results in the lab with the theoretical results that you can calculate. You can determine the % yield that you got compared with what might be expected with purely calculated values.
#35 What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper (II) sulfate? 2 Al (s) + 3 CuSO 4(aq) Al 2 (SO 4 ) 3(aq) + 3Cu (s) 1.87 g Al 1 X 1 mole Al 27.0 g Al X 3 moles Cu 2 moles Al X g Cu 1 mole Cu = 6.6 g Cu theoretical 4.65 g Cu X 100 = 70.5% 6.6 g Cu actual yield X 100 theoretical yield: Formula for % yield: