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Unit 6 – Stoichiometry and Reaction Types Cartoon courtesy of NearingZero.net.

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Presentation on theme: "Unit 6 – Stoichiometry and Reaction Types Cartoon courtesy of NearingZero.net."— Presentation transcript:

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2 Unit 6 – Stoichiometry and Reaction Types Cartoon courtesy of NearingZero.net

3 I. Types of Chemical Reactions

4 A. Combustion CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) The burning of any substance in O 2 to produce heat A + O 2  B

5 Na(s)+ O 2 (g)  C 3 H 8 (g)+ O 2 (g)  A. Combustion Products: –contain oxygen –hydrocarbons form CO 2 + H 2 O CO 2 (g)+ H 2 O(g) Na 2 O(s) 4 2

6 B. Synthesis The combination of 2 or more substances to form a compound Only one product A + B  AB

7 B. Synthesis H 2 (g) + Cl 2 (g)  2 HCl(g)

8 Al(s)+ Cl 2 (g)  AlCl 3 (s) B. Synthesis Products: –ionic - cancel charges –covalent - hard to tell

9 C. Decomposition a compound breaks down into 2 or more simpler substances only one reactant AB  A + B

10 C. Decomposition 2 H 2 O(l)  2 H 2 (g) + O 2 (g)

11 KBr(l)  K(s) + Br 2 (l) 2 C. Decomposition Products: –binary - break into elements –others - hard to tell NI 3 (s) 

12 D. Single Replacement one element replaces another in a compound –metal replaces metal (+) –nonmetal replaces nonmetal (-) A + BC  B + AC

13 D. Single Replacement Cu(s) + 2AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + 2Ag(s)

14 Fe(s)+ CuSO 4 (aq)  Cu(s)+ FeSO 4 (aq) D. Single Replacement Products: –metal  metal (+) –nonmetal  nonmetal (-) –free element must be more active (check activity series) Br 2 (l)+ NaCl(aq)  N.R.

15 Aluminum + Ferric oxide  Reaction Type: _____________________________

16 AB + CD  AD + CB E. Double Replacement ions in two compounds “change partners” cation of one compound combines with anion of the other

17 E. Double Replacement Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq)  PbCrO 4 (s) + 2KNO 3 (aq)

18 Pb(NO 3 ) 2 (aq)+ KI(aq)  PbI 2 (s)+ KNO 3 (aq) E. Double Replacement Products: –switch negative ions 2 2

19 Plumbous nitrate + Sodium iodide  Reaction Type: _____________________________ Cobaltous chloride + Sodium hydroxide Reaction Type: _____________________________

20 IQ 1 #1: Calculate the number of atoms in 47.5 g carbon. #2: Calculate the mass of 5.1 x molecules of water g C g C 1 mol C x atoms C = 2.38 x atoms C = 15 g H 2 O or 1.5 x 10 1 g H 2 O 1 mol H 2 O g H 2 O x molec H 2 O 1 mol H 2 O5.1 x molec H 2 O

21 Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet “In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.” Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

22 Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X Multiply by 6.02 X Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table

23 molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

24 Chocolate Chip Cookies!! 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

25 Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

26 Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2  2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

27 Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O –How many moles of reactants are needed? –What if we wanted 4 moles of water? –What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? –What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

28 Solving a Stoichiometry Problem 1.Balance the equation. 2.Convert masses to moles. 3.Determine which reactant is limiting. 4.Use moles of limiting reactant and mole ratios to find moles of desired product. 5.Convert from moles to grams.

29 Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

30 Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)? 2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

31 Stoichiometry Problems How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

32 Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

33 Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

34 Practice Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

35 Mass-Mole We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest –Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H O 2  4 CO H g H 2 O 1 mol H 2 O 2 mol C 2 H g H 2 O 6 mol H 2 0 = mol C 2 H 6

36 Practice Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

37 Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

38 Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N H 2  2 NH g N 2 1 mol N 2 2 mol NH g NH g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

39 IQ2 1.Determine the mass of Iron III oxide formed when 25.1 g of iron reacts with oxygen. Balanced equations: ______________________________ 2. What mass of oxygen is needed to react with 6.8 x atoms of iron? (Use the above equations) g Fe 2 O 3 1 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe 1 mol Fe g Fe 25.1 g Fe 4 Fe + 3 O 2 → 2 Fe 2 O 3 = 35.9 g Fe 2 O x atoms Fe x atoms Fe 1 mol Fe 4 mol Fe 3 mol O g O 2 1 mol O 2 = 27 g O 2

40 Example grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 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 x 2 x ÷ ÷ 4 =12.3 g Al 2 O 3

41 Example 3 How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu g Ag 1 mol Ag 12.0 g? g

42 Practice How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

43 Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

44 Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. That reactant is said to be in excess (there is too much). The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

45 Limiting Reactant To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. The lower amount of a product is the correct answer. The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

46 Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl g AlCl g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl g Cl 2 1 mol Cl 2 2 mol AlCl g AlCl g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

47 LR Example Continued We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete stop.

48 Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

49 Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

50 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

51 Limiting Reactant: Recap 1.You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2.Convert ALL of the reactants to the SAME product (pick any product you choose.) 3.The lowest answer is the correct answer. 4.The reactant that gave you the lowest answer is the LIMITING REACTANT. 5.The other reactant(s) are in EXCESS. 6.To find the amount of excess, subtract the amount used from the given amount. 7.If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!


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