Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 9 Stoichiometry Section 9.1 The Arithmetic of Equations u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a nonchemical.

Similar presentations


Presentation on theme: "Chapter 9 Stoichiometry Section 9.1 The Arithmetic of Equations u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a nonchemical."— Presentation transcript:

1

2 Chapter 9 Stoichiometry

3 Section 9.1 The Arithmetic of Equations u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a nonchemical process.

4 Section 9.1 The Arithmetic of Equations u OBJECTIVES: Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.

5 Cookies? u When baking cookies, a recipe is usually used, telling the exact amount of each ingredient If you need more, you can double or triple the amount u Thus, a recipe is much like a balanced equation

6 Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways.

7 1. In terms of Particles u Element- made of atoms u Molecular compound (made of only non- metals) = molecules u Ionic Compounds (made of a metal and non-metal parts) = formula units (ions)

8 2H 2 + O 2   2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  2 Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 2Na + 2H 2 O  2NaOH + H 2

9 Look at it differently  2H 2 + O 2   2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 x (6.02 x ) molecules of hydrogen and 1 x (6.02 x ) molecules of oxygen form 2 x (6.02 x ) molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

10 2. In terms of Moles  2 Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2 u The coefficients tell us how many moles of each substance

11 3. In terms of Mass u The Law of Conservation of Mass applies u We can check using moles  2H 2 + O 2   2H 2 O 2 moles H g H 2 1 mole H 2 =4.04 g H 2 1 mole O g O 2 1 mole O 2 =32.00 g O g H 2 +O 2

12 In terms of Mass  2H 2 + O 2   2H 2 O 2 moles H 2 O g H 2 O 1 mole H 2 O = g H 2 O 2H 2 + O 2   2H 2 O g H 2 + O 2 = g H 2 O

13 4. In terms of Volume  2H 2 + O 2   2H 2 O u At STP, 1 mol of any gas = 22.4 L  (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O) u NOTE: mass and atoms are always conserved- however, molecules, formula units, moles, and volumes will not necessarily be conserved!

14 Practice: u Show that the following equation follows the Law of Conservation of Mass: 2 Al 2 O 3  Al + 3O 2

15 Section 9.2 Chemical Calculations u OBJECTIVES: Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.

16 Section 9.2 Chemical Calculations u OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

17 Mole to Mole conversions  2 Al 2 O 3  Al + 3O 2 u each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are possible conversion factors

18 Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  2 Al 2 O 3  Al + 3O mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 =5.01 mol O 2

19 Practice: 2C 2 H O 2  4CO H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

20 How do you get good at this?

21 Mass-Mass Calculations u We do not measure moles directly, so what can we do? u We can convert grams to moles Use the Periodic Table for mass values u Then do the math with the mole ratio Balanced equation gives mole ratio! u Then turn the moles back to grams Use Periodic table values

22 For example... u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?  2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

23 More practice... u How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? Answer = 35.8 L CO 2 Answer = 58.2 L O 2 What volume of Oxygen would be required?

24 Volume-Volume Calculations u How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O2O L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH L CH 4 = 8.75 L CH L O 2 1 mol O 2 1 mol CH L CH 4

25 Avogadro told us: u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same.

26 Shortcut for Volume-Volume: u How many liters of H 2 O at STP are produced by completely burning 17.5 L of CH 4 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 2 L H 2 O = 35.0 L H 2 O Note: This only works for Volume- Volume problems.

27 Section 9.3 Limiting Reagent & Percent Yield u 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.

28 Section 9.3 Limiting Reagent & Percent Yield u OBJECTIVES: Calculate theoretical yield, actual yield, or percent yield, given appropriate information.

29 “Limiting” Reagent u If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make

30 How do you find out? u Do two stoichiometry problems. u The one that makes the least product is the limiting reagent. u For example u Copper 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?

31 u If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?  2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

32 Another example u If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? u How many grams of solid? u How much excess reagent remains?

33 Still another example u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper will be produced? u How much excess reagent will remain?

34

35 Yield u The amount of product made in a chemical reaction. u There are three types: 1. Actual yield- what you get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield 3. Percent yield = Actual Theoretical X 100

36 Example u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.  2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield?

37 Details u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %.


Download ppt "Chapter 9 Stoichiometry Section 9.1 The Arithmetic of Equations u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a nonchemical."

Similar presentations


Ads by Google