Presentation on theme: "#29 When 84.8 g of iron (III) oxide reacts with excess of carbon"— Presentation transcript:
1#29 When 84.8 g of iron (III) oxide reacts with excess of carbon monoxide, iron is produced.Fe2O3(s) CO(g) 2Fe(s) + 3CO2(g)What is the theoretical yield (how much iron do you expect to make?)of this reaction?84.8 g Fe2O311 mole Fe2O3159.7 g Fe2O32 mole Fe1mole Fe2O355.85 g Fe1mole FeXXX= g Fe
2#30 When 5.00 g copper reacts with excess silver nitrate, silver metal and copper (II) nitrate are produced. What is the theoreticalyield of silver in this reaction?Cu(s) AgNO3(aq) 2Ag(s) + Cu(NO3)2(aq)5.00 g Cu11 mole Cu63.55 g Cu2 mole Ag1 mole Cu107.9 g Ag1 mole AgXXX= g Ag
3#31 If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced.SiO2(s) C(s) SiC(s) + 2CO(g)What is the % yield in this reaction?First calculate the theoretical yield:50.0 g SiO211 mole SiO260.09 g SiO21 mole SiC1 mole SiO240.09 g SiC1 mole SiCXXX= g SiC theoretical27.9 g SiC X 100 = % yield33.36 g SiCFormula for% yield:actual yield X 100theoretical yield:
4#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 H2 2NH3calculate the theoretical yield for both nitrogen and hydrogenseparately The one that produces the least ammonia determines thetheoretical yield, then use that value to determine the % yield:15.0 g N211 mole N228 g N22 mole NH31 mole N217 g NH31 mole NH3X=XX18.2 g NH315.0 g H211 mole H22.0 g H22 mole NH33 mole H217 g NH31 mole NH3=85 g NH3XXX10.5 g NH3X 100 = %18.2 g NH3
5If a reactant runs out, the reaction must #33 In a chemical reaction, how does an insufficient quantityof a reactant affect the amount of product formed?If a reactant runs out, the reaction muststop. It is like building bicycles, if yourun out of one piece you stop buildingbicycles.
6#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 resultsthat you can calculate. You can determine the % yield that yougot compared with what might be expected with purely calculatedvalues.
7#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) CuSO4(aq) Al2(SO4)3(aq) Cu(s)1.87 g Al11 mole Al27.0 g Al3 moles Cu2 moles Alg Cu1 mole CuXXX= 6.6 g Cutheoretical4.65 g Cu X = 70.5%6.6 g CuFormula for% yield:actual yield X 100theoretical yield: