10 C. Decomposition 2 2 K(s) + Br2(l) KBr(l) NI3(s) Products: binary - break into elementsothers - hard to tellK(s) + Br2(l)KBr(l) NI3(s)
11 A + BC B + AC D. Single Replacement one element replaces another in a compoundmetal replaces metal (+)nonmetal replaces nonmetal (-)A + BC B + AC
12 Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) D. Single ReplacementCu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)
13 D. Single Replacement Fe(s)+ CuSO4(aq) Cu(s)+ FeSO4(aq) Products:metal metal (+)nonmetal nonmetal (-)free element must be more active (check activity series)Fe(s)+ CuSO4(aq) Cu(s)+ FeSO4(aq)Br2(l)+ NaCl(aq) N.R.
19 IQ 1 #1: Calculate the number of atoms in 47.5 g carbon. #2: Calculate the mass of 5.1 x 1023 molecules of water.47.5 g C1 mol C6.022 x 1023 atoms C12.01 g C1 mol C= x atoms C5.1 x 1023 molec H2O1 mol H2O18.02 g H2O1 mol H2O6.022 x 1023 molec H2O= 15 g H2O or 1.5 x 101 g H2O
20 Stoichiometry“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.”Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in ScarletStoichiometry - The study of quantities of materials consumed and produced in chemical reactions.
21 Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 1023 Multiply by 6.02 X 1023MolesMultiply by atomic/molar mass from periodic tableDivide by atomic/molar mass from periodic tableMass (grams)
22 Everything must go through Moles!!! Calculationsmolar mass Avogadro’s number Grams Moles particlesEverything must go through Moles!!!
23 Chocolate Chip Cookies!! 1 cup butter1/2 cup white sugar1 cup packed brown sugar1 teaspoon vanilla extract2 eggs2 1/2 cups all-purpose flour1 teaspoon baking soda1 teaspoon salt2 cups semisweet chocolate chipsMakes 3 dozenHow 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?
24 Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as wellInstead of calling them recipes, we call them reaction equationsFurthermore, instead of using cups and teaspoons, we use molesLastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients
25 Chemistry RecipesLooking 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 + Cl2 2 NaClThis reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chlorideWhat if we wanted 4 moles of NaCl? 10 moles? 50 moles?
26 PracticeWrite the balanced reaction for hydrogen gas reacting with oxygen gas.2 H2 + O2 2 H2OHow 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?
27 Solving a Stoichiometry Problem Balance the equation.Convert masses to moles.Determine which reactant is limiting.Use moles of limiting reactant and mole ratios to find moles of desired product.Convert from moles to grams.
28 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.4Al+3O22Al2O3a. Every reaction needs a yield sign!b. What are the reactants?c. What are the products?d. What are the balanced coefficients?
29 Mole RatiosThese mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemicalExample: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?2 Na + Cl2 2 NaCl5 moles Na 1 mol Cl22 mol Na= 2.5 moles Cl2
30 Stoichiometry Problems How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?2KClO3 2KCl + 3O2? mol9 mol9 mol O22 mol KClO33 mol O2= 6 molKClO3
31 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?
32 Mole-Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of molesWe still go through moles and use the mole ratio, but now we also use molar mass to get to gramsExample: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?2 Na + Cl2 2 NaCl5.00 moles Na 1 mol Cl g Cl22 mol Na 1 mol Cl2= 177g Cl2
33 PracticeCalculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.
34 Mass-MoleWe can also start with mass and convert to moles of product or another reactantWe use molar mass and the mole ratio to get to moles of the compound of interestCalculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water2 C2H6 + 7 O2 4 CO2 + 6 H2010.0 g H2O 1 mol H2O 2 mol C2H618.0 g H2O 6 mol H20= mol C2H6
35 PracticeCalculate how many moles of oxygen are required to make 10.0 g of aluminum oxide
36 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
37 Mass-Mass ConversionEx. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.N2 + 3 H2 2 NH32.00g N mol N mol NH g NH328.02g N2 1 mol N mol NH3= 2.4 g NH3
38 IQ2Determine 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 1023 atoms of iron? (Use the above equations).4 Fe + 3 O2 → 2 Fe2O325.1 g Fe2 mol Fe2O3g Fe2O31 mol Fe55.85 g Fe4 mol Fe1 mol Fe2O3= 35.9 g Fe2O36.8 x 1023 atoms Fe1 mol Fe3 mol O232.00 g O26.022 x 1023 atoms Fe4 mol Fe1 mol O2= 27 g O2
39 Example 26.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?4 Al O2 2Al2O36.50 g Al1 mol Al2 mol Al2O3g Al2O3=? g Al2O326.98 g Al4 mol Al1 mol Al2O36.50 x 2 x ÷ ÷ 4 =12.3 g Al2O3
40 Example 3 Cu + 2AgNO3 2Ag + Cu(NO3)2 12.0 g Cu 1 mol Cu 63.55 g Cu How many grams of silver will be formed from 12.0 g copper?Cu + 2AgNO3 2Ag + Cu(NO3)212.0 g? g12.0g Cu1 molCu63.55g Cu2 molAg1 molCu107.87g Ag1 molAg= 40.7 gAg
41 PracticeHow many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?
42 Limiting Reactant: Cookies 1 cup butter1/2 cup white sugar1 cup packed brown sugar1 teaspoon vanilla extract2 eggs2 1/2 cups all-purpose flour1 teaspoon baking soda1 teaspoon salt2 cups semisweet chocolate chipsMakes 3 dozenIf 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?
43 Limiting ReactantMost 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.
44 Limiting ReactantTo 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!
45 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 Cl2 2 AlCl3Start with Al:Now Cl2:10.0 g Al 1 mol Al mol AlCl g AlCl327.0 g Al mol Al mol AlCl3= 49.4g AlCl335.0g Cl mol Cl mol AlCl g AlCl371.0 g Cl mol Cl mol AlCl3= 43.9g AlCl3
46 LR Example ContinuedWe 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 .
47 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.
48 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?
49 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KIWe found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.15.0 g I mol I mol K g K254 g I mol I mol K= 4.62 g K USED!15.0 g K – 4.62 g K = g K EXCESSNote that we started with the limiting reactant! Once you determine the LR, you should only start with it!Given amount of excess reactantAmount of excess reactant actually used
50 Limiting Reactant: Recap You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.Convert ALL of the reactants to the SAME product (pick any product you choose.)The lowest answer is the correct answer.The reactant that gave you the lowest answer is the LIMITING REACTANT.The other reactant(s) are in EXCESS.To find the amount of excess, subtract the amount used from the given amount.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!