2Section 8.1 The Arithmetic of Equations OBJECTIVES:Calculate the amount of reactants required, or product formed, in a non chemical process.
3Section 8.1 The Arithmetic of Equations OBJECTIVES:Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.
4Cookies?When baking cookies, a recipe is usually used, telling the exact amount of each ingredientIf you need more, you can double or triple the amountThus, a recipe is much like a balanced equation
5Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation.We can interpret balanced chemical equations several ways.
61. In terms of Particles Element- made of atoms Molecular compound (made of only non- metals) = moleculesIonic Compounds (made of a metal and non-metal parts) = formula units (ions)
72H2 + O2 ® 2H2OTwo molecules of hydrogen and one molecule of oxygen form two molecules of water.2 Al2O3 ® 4Al + 3O22formula unitsAl2O3form4atomsAland3moleculesO22Na + 2H2O ® 2NaOH + H2
8Look at it differently 2H2 + O2 ® 2H2O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water.2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water.2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.
92. In terms of Moles 2 Al2O3 ® 4Al + 3O2 2Na + 2H2O ® 2NaOH + H2 The coefficients tell us how many moles of each substance
103. In terms of MassThe Law of Conservation of Mass appliesWe can check using moles2H2 + O2 ® 2H2O2.02 g H22 moles H2=4.04 g H21 mole H232.00 g O21 mole O2=32.00 g O21 mole O236.04 g H2+O236.04 g H2+O2
11In terms of Mass 2H2 + O2 ® 2H2O 2H2 + O2 ® 2H2O 18.02 g H2O 2 moles H2O=1 mole H2O2H2 + O2 ® 2H2O36.04 g H2 + O2=36.04 g H2O
124. In terms of Volume 2H2 + O2 ® 2H2O At STP, 1 mol of any gas = 22.4 L(2 x 22.4 L H2) + (1 x 22.4 L O2) ® (2 x 22.4 L H2O)NOTE: mass and atoms are always conserved- however, molecules, formula units, moles, and volumes will not necessarily be conserved!
13Practice:Show that the following equation follows the Law of Conservation of Mass:2 Al2O3 ® 4Al + 3O2
14Section 8.2 Chemical Calculations OBJECTIVES:Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.
15Section 8.2 Chemical Calculations OBJECTIVES:Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.
16Mole to Mole conversions 2 Al2O3 ® 4Al + 3O2each time we use 2 moles of Al2O3 we will also make 3 moles of O22 moles Al2O33 mole O2or3 mole O22 moles Al2O3These are possible conversion factors
17Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?2 Al2O3 ® 4Al + 3O23 mol O23.34 mol Al2O3=5.01 mol O22 mol Al2O3
18Practice:2C2H2 + 5 O2 ® 4CO2 + 2 H2OIf 3.84 moles of C2H2 are burned, how many moles of O2 are needed?(9.6 mol)How many moles of C2H2 are needed to produce 8.95 mole of H2O?(8.95 mol)If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?(4.94 mol)
20Mass-Mass Calculations We do not measure moles directly, so what can we do?We can convert grams to molesUse the Periodic Table for mass valuesThen do the math with the mole ratioBalanced equation gives mole ratio!Then turn the moles back to gramsUse Periodic table values
21For example...If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?2Fe + 3CuSO4 ® Fe2(SO4)3 + 3CuAnswer = 17.2 g Cu
22More practice...How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ?Answer = 35.8 L CO2What volume of Oxygen would be required?Answer = 58.2 L O2
23Volume-Volume Calculations How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ?CH4 + 2O2 ® CO2 + 2H2O22.4 L O21 mol O21 mol CH41 mol CH422.4 L CH41 mol O222.4 L CH417.5 L O222.4 L O22 mol O21 mol CH4= 8.75 L CH4
24Avogadro told us:Equal volumes of gas, at the same temperature and pressure contain the same number of particles.Moles are numbers of particlesYou can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.
25Shortcut for Volume-Volume: How many liters of H2O at STP are produced by completely burning L of CH4 ?CH4 + 2O2 ® CO2 + 2H2O2 L H2O17.5 L CH4= 35.0 L H2O1 L CH4Note: This only works for Volume-Volume problems.
26Section 8.3 Limiting Reagent & Percent Yield OBJECTIVES:Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced, and the amount of excess reagent.
27Section 8.3 Limiting Reagent & Percent Yield OBJECTIVES:Calculate theoretical yield, actual yield, or percent yield, given appropriate information.
28“Limiting” ReagentIf you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make?The limiting reagent is the reactant you run out of first.The excess reagent is the one you have left over.The limiting reagent determines how much product you can make
29How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent.For exampleCopper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?
30Cu is the Limiting Reagent because it creates less product If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?2Cu + S ® Cu2SCu is the Limiting Reagent because it creates less product1 mol Cu1 mol Cu2Sg Cu2S10.6 g Cu63.55g Cu2 mol Cu1 mol Cu2S= 13.3 g Cu2S= 13.3 g Cu2S1 mol S1 mol Cu2Sg Cu2S3.83 g S32.06g S1 mol S1 mol Cu2S= 19.0 g Cu2S
31Another exampleIf 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?How many grams of solid?How much excess reagent remains?Limiting Reagent Simulation
32Still another exampleIf 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?How much excess reagent will remain?
34Yield The amount of product made in a chemical reaction. There are three types:1. Actual yield- what you get in the lab when the chemicals are mixed2. Theoretical yield- what the balanced equation tells should be made3. Percent yield = ActualTheoreticalX 100
35Example6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.2Al + 3 CuSO4 ® Al2(SO4)3 + 3CuWhat is the actual yield?What is the theoretical yield?What is the percent yield?
36% Yield Problem AY is 6.78 g of Cu TY – Use the balanced equation and do a mass to mass calculation2Al + 3 CuSO4 Al2(SO4)3 + 3Cu3.92 gx g3.92 g Al1 mol Al3mol Cu63.55 g Cu13.9 g Cu26.98 g Al2 mol Al1 mol Cu% Yield = (AY / TY) x 100 = (6.78g / g) x 100 = % yield
37Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %.
38The Heat of ReactionAt the start of a reaction the reactants have a great heat content.Enthalpy is the amount of heat that a substance has at a giventemperature and pressure (see Table 8.1 pg 190)The heat of a reaction is the heat that is released or absorbedduring a chemical reaction. Heat of Reaction is represented byThe symbol H
39Heat can be added to make a reaction occur. This is an Endothermic Reaction H will be positive. The reaction can be represented asfollows:2 NH N H H = kJor2 NH kJ N H2
41When heat is released by a chemical reaction, the reaction is said to be Exothermic. H will be negativeHeat is most often released by a chemical reaction. The releasedheat can be represented by the following:Na + Cl NaCl H = kJorNa + Cl NaCl kJHow can you calculate the heat of a reaction???
43Calculate the heat of reaction ( H), in kilojoules, for the reaction Use table 8.1 pg 1904 CO2(g) O2(g) CO(g)4(-393.5)2(0)+4(-110.5)(-1574)(0)+(-442)H = NRG of Products – NRG of ReactantsH = (-442 kJ) – (-1574)H = 1132 kJ(+ indicates an endothermic rxn so)4 CO2(g) kJ O2(g) CO(g)How much energy is used to make 100g of CO(g)?
444 CO2(g) kJ O2(g) CO(g)x kJ100g100 g CO1 mol CO1132 kJ1010 kJ4 mol CO28.01 g CO