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Chapter 11 “Stoichiometry” Mr. Mole. Let’s make some Cookies! When baking cookies, a recipe is usually used, telling the exact amount of each ingredient.

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Presentation on theme: "Chapter 11 “Stoichiometry” Mr. Mole. Let’s make some Cookies! When baking cookies, a recipe is usually used, telling the exact amount of each ingredient."— Presentation transcript:

1 Chapter 11 “Stoichiometry” Mr. Mole

2 Let’s make some Cookies! 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 Thus, a recipe is much like a balanced equation.

3 Stoichiometry is… Greek for “measuring elements” Pronounced “stoy kee ahm uh tree” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are 3 ways to interpret a balanced chemical equation

4 #1. In terms of Particles An Element is made of atoms A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements) Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

5 Example: 2H 2 + O 2 → 2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Another example: 2Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

6 #2. In terms of Moles The coefficients tell us how many moles of each substance 2Al 2 O 3  Al + 3O 2 2Na + 2H 2 O  2NaOH + H 2 Remember: A balanced equation is a Molar Ratio

7 #3. In terms of Mass The Law of Conservation of Mass applies We can check mass by 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 = g O g H 2 + O 2 + reactants

8 In terms of Mass (for products) 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 g H 2 + O 2 = g H 2 O The mass of the reactants must equal the mass of the products grams reactant = grams product

9 Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al 2 O 3  Al + 3O 2

10 Practice p 371 1).Interpret these equations in terms of particles, moles and mass. Show that the law of conservation is observed. -N 2 (g)+3H 2 (g)--> 2 NH 3 (g) -HCl(aq)+KOH(aq)-->KCl(aq)+H 2 O(l) -2Mg(s)+O 2 (g)-->2MgO(s) 2).Balance each equation and interpret in terms of particles, moles and mass. Also show that the law of conservation is observed. __Na(s)+__H 2 0(l)-->__NaOH(aq)+__H 2 (g) __Zn(s)+__HNO 3 (aq)--> Zn(NO 3 ) 2 (aq)+__N 2 O(g)+__H 2 O(l)

11 Practice p 372 3). Determine all possible mole ratios a. 4Al+3O 2 -->2Al 2 O 3 b. 3Fe+4H 2 O-->Fe 3 O 4 +4H 2 c. 2H 2 O-->2H 2 +O 2 4). Balance the following equations and determine the possible mole ratios- a. ZnO+HCl-->ZnCl 2 +H 2 O b. C 4 H 10 +oxygen-->carbon dioxide+water

12 Mole to Mole conversions 2Al 2 O 3  Al + 3O 2 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 the two possible conversion factors to use in the solution of the problem.

13 Mole to Mole conversions How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 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 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

14 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)

15 How do you get good at this?

16 Practice p Methane and sulfur react to produce CS 2. __CH 4 +__S 8 -->__CS 2 +__H 2 S a.Balance the equation. b. Caculate the moles of CS 2 produced when 1.5 moles S 8 is used. c. How many moles of H 2 S are produced? 12. Sulfuric acid, H 2 SO 4, is formed when sulfur dioxide SO 2 reacts with oxygen and water. a. Write the balanced chemical equation for the reaction. b. How many moles of H 2 SO 4 are produced from 12.5 moles of SO 2 ? c. How many moles of O 2 are needed?

17 Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.

18 Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O g Al 2 O 3 (6.50 x 1 x 2 x ) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

19 Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

20 Practice p ). 2.5 moles of NaCl are decomposed by passing electricity. How much Cl 2 (in g) is obtained? 14). TiO 2 +C+2Cl 2 --> TiCl 4 +CO 2 a. What mass of Cl 2 gas is needed to react with 1.25 mol of TiO 2 ? b. What mass of C is needed to react with 1.25 mol of TiO 2 ? c. What is the mass of all the products formed by reaction with 1.25 mol of TiO 2 ?

21 Practice p ). In the reaction, 2NaN 3 -->2Na+3N 2, determine the mass of N 2 produced from 100 g of NaN 3 ? 16). In the formation of acid rain, SO 2 reacts with oxygen and water to form sulfuric acid H 2 SO 4. Write the balanced equation for the reaction. If 2.5 g of SO 2 reacts with excess oxygen and water, how much H 2 SO 4, in grams, is produced?

22 Volume-Volume Calculations: 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 O 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 Notice anything relating these two steps?

23 Avogadro told us: Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 1 mole = 22.4 STP

24 Shortcut for Volume-Volume? 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 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note: This only works for Volume-Volume problems.

25 “Limiting” Reagent 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? 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

26 Limiting Reagents - Combustion

27 How do you find out which is limited? The chemical that makes the least amount of product is the “limiting reagent”. You can recognize limiting reagent problems because they will give you 2 amounts of chemical Do two stoichiometry problems, one for each reagent you are given.

28 If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) 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 the Limiting Reagent, since it produced less product.

29 Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 the CuSO 4 is limited, so Cu = 20.6 g How much excess reagent will remain? Excess = 4.47 grams

30 Practice p Na+Fe 2 O 3 -->3Na 2 O+2Fe If 100g of Na and 100g of Fe 2 O 3 are used in this reaction, determine the following- limiting reactant, reactant in excess, mass of solid iron produced, mass of excess reactant that remains after reaction is completed. 24. A plant has 88g of CO 2 and 64g of H 2 O to produce glucose, C 6 H 12 O 6, by photosynthesis. a. Write the balanced equation for it. b. Determine the limiting reagent. c. Determine the excess reagent. d. Determine the mass in excess. e. Determine the mass of glucose produced.

31 p394 Q 81- Zn+2 MnO 2 +H 2 O-->Zn(OH) 2 +Mn 2 O 3 Determine the LR if 25 g of Zn and 30g of MnO2 are used. Determine the mass of Zn(OH) 2 produced.

32 The Concept of: A little different type of yield than you had in Driver’s Education class.

33 What is Yield? Yield is the amount of product made in a chemical reaction. There are three types: 1. Actual yield- what you actually 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

34 Example: 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 What is the actual yield? What is the theoretical yield? What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

35 Details on Yield Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring

36 Practice p Al(OH) 3 +3HCl-->AlCl 3 +3H 2 O, If 14g of Al(OH) 3 is present, determine the theoretical yield of AlCl 3 produced when the tablet reacts with HCl. 29. Zn+I 2 -->ZnI 2 (a).Determine the theoretical yield if mol of zinc is used. (b).Determine the percent yield is 515.6g of product is recovered. 30. When copper wire is placed into a silver nitrate AgNO 3 solution, silver crystals and Cu(NO 3 ) 2 solution form. a. Write the balanced chemical equation for the reaction. b. If a 20g sample of copper is used, determine the theoretical yield of silver. c. If 60g of silver is recovered from the reaction, determine the percent yield of the reaction.

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