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Unit 11 Stoichiometry CP Chemistry Big Ideas u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a chemical reaction. Interpret.

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Presentation on theme: "Unit 11 Stoichiometry CP Chemistry Big Ideas u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a chemical reaction. Interpret."— Presentation transcript:

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2 Unit 11 Stoichiometry CP Chemistry

3 Big Ideas u OBJECTIVES: Calculate the amount of reactants required, or product formed, in a chemical reaction. Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.

4 Remember The Mole u 1 mole = 6.02 X 10 23 particles u 1 mole = the atomic weight in grams of the element u 1 mole = the atomic weight in grams of the compound

5 What is a Particle? u Element Made of atoms u Molecules Molecular compound 2 non-metals u Formula Units Ionic Compounds Metal and non-metal

6 Let’s Review u How many formula units are in 2.75 moles of CaCl 2 ? 1.66x10 24 formula units u How many total atoms are in 2.75 moles of CaCl 2 ? 4.97x10 24 atoms

7 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

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

9 In terms of Particles  2H 2 + O 2   2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  Al 2 S 3  Al + 3S 1 formula unit of Al 2 S 3 decomposes to 2 atoms of Al + 3 atoms of S  2Na + 2H 2 O  2NaOH + H 2

10 In terms of Moles u The coefficients tell us how many moles of each substance  2 Al 2 O 3  Al + 3O 2 2 moles of Al 2 O 3 decompose to 4 moles Al + 3 moles O 2  2Na + 2H 2 O  2NaOH + H 2

11 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 2 2.02 g H 2 1 mole H 2 =4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 =32.00 g O 2 36.04 g H 2 +O 2

12 In terms of Mass  2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 2H 2 + O 2   2H 2 O 36.04 g H 2 + O 2 = 36.04 g H 2 O

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

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

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

16 Mole to Mole conversions  2 Al 2 O 3  Al + 3O 2 Each time we use 2 moles of Al 2 O 3 we make 3 moles of O 2 Possible conversion factors: 2 moles Al 2 O 3 3 mole O 2 or 3 moles O 2 2 moles Al 2 O 3

17 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 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 =5.01 mol O 2

18 Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.60 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)

19 How do you get good at this?

20 Mass Stoichiometry Calculations u We do not measure moles directly, so what can we do? u We can convert between grams and moles Use Periodic Table for molar mass

21 EXTREMELY IMPORTANT!!! u Label what substance your factor is for 1 mol 1 mol NaCl

22 Mass to Mole Calculations u 2 step conversions: u 1 st step: mass to moles Divide by molar mass of what’s given u 2 nd step: moles of one to moles of another Use mole ratio

23 Mass to Mole Calculations 2Na + 2H 2 O  2NaOH + H 2 u If 23.5g of NaOH react, how many moles of H 2 are produced? u 23.5g NaOH u 0.294 mol H 2 40.00g NaOH 1 mol NaOH 2 mol NaOH 1 mol H2H2

24 Mole to Mass Calculations u 2 step conversions: u 1 st step: moles of one to moles of another Use mole ratio u 2 nd step: moles to mass Multiply by molar mass of new substance

25 Mole to Mass Calculations 2Na + 2H 2 O  2NaOH + H 2 u If 3.47 moles of H 2 are produced, what mass of NaOH reacted? u 3.47 mol H 2 u 278g NaOH 1 mol H2H2 2 NaOH 1 mol NaOH 40.00g NaOH

26 Mass to Mass Conversions u 1 st step: convert grams to moles Divide by molar mass of what’s given u 2 nd step: moles of one to moles of the other Use mole ratio u 3 rd step: Convert the moles back to grams Multiply by molar mass of new substance

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

28 Mass to Mass Conversions 2 Al 2 O 3  Al + 3O 2 u What mass of aluminum can be formed from 22.8g of aluminum oxide? Answer: 12.1g Al u If 3.47g of O 2 are produced, what mass of Al 2 O 3 reacted? Answer: 7.37g Al 2 O 3

29 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.

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

31 “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

32 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: 2Cu + S  Cu 2 S If 10.6 g of copper reacts with 3.83 g S how much copper (II) sulfide will be formed?

33 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 159.16 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 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is Limiting Reagent

34 Another Example 2Al + 3CuSO 4  Al 2 (SO 4 ) 3 + 3Cu 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?

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36 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

37 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?

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

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40 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!

41 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 O2O2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

42 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.

43 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.

44 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?

45 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?


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