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Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 10 Chemical Equation Calculations INTRODUCTORY CHEMISTRY INTRODUCTORY.

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Presentation on theme: "Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 10 Chemical Equation Calculations INTRODUCTORY CHEMISTRY INTRODUCTORY."— Presentation transcript:

1 Christopher G. Hamaker, Illinois State University, Normal IL © 2008, Prentice Hall Chapter 10 Chemical Equation Calculations INTRODUCTORY CHEMISTRY INTRODUCTORY CHEMISTRY Concepts & Connections Fifth Edition by Charles H. Corwin

2 Chapter 10 2 Chemists and chemical engineers must perform calculations based on balanced chemical reactions to predict the cost of processes. These calculations are used to avoid using large, excess amounts of costly chemicals. The calculations these scientists use are called stoichiometry calculations. What is Stoichiometry?

3 Chapter 10 3 Let’s look at the reaction of nitrogen monoxide with oxygen to produce nitrogen dioxide: 2 NO(g) + O 2 (g) → 2 NO 2 (g) Two molecules of NO gas react with one molecule of O 2 gas to produce 2 molecules of NO 2 gas. You can see from the following presentation that during a chemical reaction, the atoms are neither destroyed nor created but simply re-arranged. Interpreting Chemical Equations

4 Chapter NO(g) + O 2 (g) → 2 NO 2 (g) The coefficients represent molecules (or moles), so we can multiply each of the coefficients and look at more than individual molecules. NO(g)O2(g)O2(g)NO 2 (g) *2 molecules1 molecule2 molecules *2000 molecules1000 molecules2000 molecules *12.04 × molecules 6.02 × molecules × molecules *2 moles1 mole2 moles Moles & Equation Coefficients

5 Chapter NO(g) + O 2 (g) → 2 NO 2 (g) We can now read the balanced chemical equation as “2 moles of NO gas react with 1 mole of O 2 gas to produce 2 moles of NO 2 gas.” ***The coefficients indicate the ratio of moles, or mole ratio, of reactants and products in every balanced chemical equation.*** Mole Ratios

6 Chapter 10 6 Recall that, according to Avogadro’s theory, there are equal number of molecules in equal volumes of gas at the same temperature and pressure. So, twice the number of molecules occupies twice the volume. 2 NO(g) + O 2 (g) → 2 NO 2 (g) So, instead of 2 molecules NO, 1 molecule O 2, and 2 molecules NO 2, we can write: 2 liters of NO react with 1 liter of O 2 gas to produce 2 liters of NO 2 gas. Volume & Equation Coefficients

7 Chapter 10 7 From a balanced chemical equation, we know how many molecules or moles of a substance react and how many moles of product(s) are produced. If there are gases, we know how many liters of gas react or are produced. Interpretation of Coefficients

8 Chapter 10 8 The law of conservation of mass states that mass (or atom) is neither created nor destroyed during a chemical reaction. Let’s test: 2 NO(g) + O 2 (g) → 2 NO 2 (g) –2 mol NO + 1 mol O 2 → 2 mol NO 2 –2 (30.01 g) + 1 (32.00 g) → 2 (46.01 g) –60.02 g g → g –92.02 g = g The mass of the reactants is equal to the mass of the product! Mass is conserved. Conservation of Mass

9 Chapter 10 9 We can use a balanced chemical equation to write mole ratio, which can be used as unit (or say conversion) factors: N 2 (g) + O 2 (g) → 2 NO(g) Since 1 mol of N 2 reacts with 1 mol of O 2 to produce 2 mol of NO, we can write the following mole relationships: 1 mol N 2 1 mol O 2 1 mol N 2 2 mol NO 1 mol O 2 2 mol NO 1 mol O 2 1 mol N 2 2 mol NO 1 mol N 2 2 mol NO 1 mol O 2 Mole-Mole Relationships

10 Chapter How many moles of oxygen react with 2.25 mol of nitrogen? N 2 (g) + O 2 (g) → 2 NO(g) We want mol O 2 (substance B); we have 2.25 mol N 2 (substance A) ***Here the quantity (gram or mole) of substance A is always given, while the quantity of substance is always asked. The method used in the following is called cancelling method in Nutrition class and is called dimensional analysis in General Chemistry class. We always use the unit factor as desired unit/given unit.*** Use 1 mol N 2 = 1 mol O 2. = 2.25 mol O mol N 2 × 1 mol O 2 1 mol N 2 Mole-Mole Calculations

11 Chapter Critical Thinking: Steelmaking What is the difference between iron and steel? Iron is the pure element Fe. Steel is an alloy of iron with other elements. –Other elements are included in steel to impart special properties, such as increased strength or resistance to corrosion. –Common additive elements in steel include carbon, manganese, and chromium. –**Alloy is a solid-type homogeneous mixture.**

12 Chapter There are three basic types of stoichiometry problems we’ll introduce in this chapter: –Mass-Mass stoichiometry problems –Mass-Volume stoichiometry problems –Volume-Volume stoichiometry problems Types of Stoichiometry Problems

13 Chapter In a mass-mass stoichiometry problem, we will convert a given mass of a reactant or product to an unknown mass of reactant or product. There are three steps: –Convert the given mass of substance (i.e. A) to moles using the molar mass of the substance as a unit factor. mole of A = mass of A ÷ molar mass of A –Convert the moles of the given to moles of the unknown using the coefficients in the balanced equation. mole of B = mole of A × (coefficient of B ÷ coefficient of A) –Convert the moles of the unknown (i.e. B) to grams using the molar mass of the substance as a unit factor. mass of B = mole of B × molar mass of B *****Mass-Mass Problems*****

14 Chapter What is the mass of mercury produced from the decomposition of 1.25 g of orange mercury(II) oxide (MM = g/mol)? 2 HgO(s) → 2 Hg(l) + O 2 (g) Convert grams HgO to moles HgO using the molar mass of mercury ( g/mol): Mole HgO = mass HgO ÷ molar mass HgO = 1.25 g ÷ g/mol = Convert moles Hg to moles HgO using the balanced equation mol HgO × (2 Hg ÷ 2 HgO) = mol Hg Convert moles Hg to grams Hg using the molar mass mol Hg × g/mol Hg = 1.16 grams Hg *****Mass-Mass Problem*****

15 Chapter HgO(s) → 2 Hg(l) + O 2 (g) g Hg  mol Hg  mol HgO  g HgO Solve by unit factors ( i.e. cancelling method or dimensional analysis); Convert this to Road Map method manually on the board. = 1.16 g Hg 1.25 g HgO × 2 mol Hg 2 mol HgO 1 mol HgO g HgO × 1 mol Hg g Hg × *****Problem, continued*****

16 Chapter In a mass-volume stoichiometry problem, we will convert a given mass of a reactant or product to an unknown volume of reactant or product. There are three steps: –Convert the given mass of a substance to moles using the molar mass of the substance as a unit factor. –Convert the moles of the given to moles of the unknown using the coefficients in the balanced equation. –Convert the moles of unknown to liters using the molar volume of a gas as a unit factor. Mass-Volume Problems

17 Chapter How many liters of hydrogen are produced from the reaction of g of aluminum metal with dilute hydrochloric acid? 2 Al(s) + 6 HCl(aq) → 2 AlCl 3 (aq) + 3 H 2 (g) Convert grams Al to moles Al using the molar mass of aluminum (26.98 g/mol). mole Al = g ÷ g/mol = mol Convert moles Al to moles H 2 using the balanced equation. mol H 2 = mol Al × {3 H 2 / 2 Al} = mol Convert moles H 2 to liters using the molar volume at STP. STP refers to standard temperature and pressure which means 0 o C and 1 atm. liters H 2 = mole H 2 × 22.4 liters/mole = × 22.4 = *****Mass-Volume Problem*****

18 Chapter Al(s) + 6 HCl(aq) → 2 AlCl 3 (aq) + 3 H 2 (g) g Al  mol Al  mol H 2  L H 2 = L H g Al × 3 mol H 2 2 mol Al 1 mol Al g Al × 1 mol H L H 2 × Problem, continued

19 Chapter *****Problem, continued***** Another way to work on this problem is to use ideal gas equation, PV = nRT to figure out the volume of hydrogen gas. Here, P is the pressure in atm, V is the volume in liter, n is the mole of the gas, T is the temperature in Kelvin, and R is the ideal gas constant and its value is atm.l/mole.k. STP means 0 o C and 1 atm. PV = nRT  1 x V = x x ( ) So V = L.

20 Chapter How many grams of sodium chlorate are needed to produce 9.21 L of oxygen gas at STP? 2 NaClO 3 (s) → 2 NaCl(s) + 3 O 2 (g) Convert liters of O 2 to moles O 2, to moles NaClO 3, to grams NaClO 3 ( g/mol). Volume-Mass Problem × 1 mol NaClO g NaClO 3 × 9.21 L O 2 × 1 mol O L O 2 2 mol NaClO 3 3 mol O 2 = 29.2 g NaClO 3

21 Chapter *****Problem, Continued***** Another way (road map method) to solve this problem is as following: How many grams of sodium chlorate are needed to produce 9.21 L of oxygen gas at STP? Molar mass of NaClO 3 is g/mol. 2 NaClO 3 (s) → 2 NaCl(s) + 3 O 2 (g) Substance A is O 2 as the quantity is given as 9.21 L at STP (1 atm, 0 o C). Actually the mole of oxygen gas is indirectly given: From PV = nRT  1 × 9.21 = n × ×  n = 9.21 ÷ = Substance B is NaClO 3 as it’s asked. From the coefficient information: mole of NaClO 3 = mole O 2 × {2 NaClO 3 /3 O 2 (g)} = So mass of NaClO 3 = × = g, round to 29.2 g.

22 Chapter Gay-Lussac discovered that volumes of gases under similar conditions combined in small whole number ratios. This is the law of combining volumes. Consider the reaction: H 2 (g) + Cl 2 (g) → 2 HCl(g) 10 mL of H 2 reacts with 10 mL of Cl 2 to produce 20 mL of HCl. The ratio of volumes is 1:1:2, small whole numbers. Volume-Volume Stoichiometry

23 Chapter The whole number ratio (1:1:2) is the same as the mole ratio in the balanced chemical equation: H 2 (g) + Cl 2 (g) → 2 HCl(g) Law of Combining Volumes

24 Chapter In a volume-volume stoichiometry problem, we will convert a given volume of a gas to an unknown volume of gaseous reactant or product. There is one step: –Convert the given volume to the unknown volume using the mole ratio (therefore, the volume ratio) from the balanced chemical equation. Volume-Volume Problems

25 Chapter How many liters of oxygen react with 37.5 L of sulfur dioxide in the production of sulfur trioxide gas? 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) From the balanced equation, 1 mol of oxygen reacts with 2 mol sulfur dioxide. So, 1 L of O 2 reacts with 2 L of SO 2. Volume-Volume Problem

26 Chapter SO 2 (g) + O 2 (g) → 2 SO 3 (g) L SO 2  L O 2 = 18.8 L O L SO 2 × 1 L O 2 2 L SO 2 = 37.5 L SO L SO 2 × 2 L SO 3 2 L SO 2 How many L of SO 3 are produced? Problem, continued

27 Chapter Chemistry Connection: Ammonia Ammonia, the common household cleaner, is one of the 10 most important industrial chemicals. Household cleaning uses only a small portion of the ammonia produced. Ammonia is very important as a fertilizer in agriculture. Nitrogen is an essential nutrient for plants, but most cannot use atmospheric N 2.

28 Chapter Say you’re making grilled cheese sandwiches. You need 1 slice of cheese and 2 slices of bread to make one sandwich. 1 Cheese + 2 Bread → 1 Sandwich If you have 5 slices of cheese and 8 slices of bread, how many sandwiches can you make? You have enough bread for 4 sandwiches and enough cheese for 5 sandwiches. You can only make 4 sandwiches; you will run out of bread before you use all the cheese. *****Limiting Reactant Concept*****

29 Chapter Since you run out of bread first, bread is the ingredient that limits how many sandwiches you can make. In a chemical reaction, the limiting reactant is the reactant that controls the amount of product you can make. A limiting reactant is used up before the other reactants. The other reactants are present in excess and are called excess reactants. Limiting Reactant

30 Chapter If you heat 2.50 mol of Fe and 3.00 mol of S, how many moles of FeS are formed? Fe(s) + S(s) → FeS(s) According to the balanced equation, 1 mol of Fe reacts with 1 mol of S to give 1 mol of FeS. So 2.50 mol of Fe will react with 2.50 mol of S to produce 2.50 mol of FeS. Therefore, iron is the limiting reactant and sulfur is the excess reactant. *****Determining the Limiting Reactant*****

31 Chapter If you start with 3.00 mol of sulfur and 2.50 mol of sulfur reacts to produce FeS, you have 0.50 mol of excess sulfur (3.00 mol – 2.50 mol). The table below summarizes the amounts of each substance before and after the reaction. Determining the Limiting Reactant

32 Chapter There are three steps to a limiting reactant problem: 1.Calculate the mass of product that can be produced from the first reactant. mass reactant #1  mol reactant #1  mol product  mass product 2.Calculate the mass of product that can be produced from the second reactant. mass reactant #2  mol reactant #2  mol product  mass product 3.The limiting reactant is the reactant that produces the least amount of product. 4.There are two ways to solve the limiting reactant problem: (1) the reactant-approaching method, not shown in the textbook and (2) the product-approaching method, shown in textbook and is the recommended one. *****Mass Limiting Reactant Problems*****

33 Chapter How much molten iron is formed from the reaction of 25.0 g FeO and 25.0 g Al? 3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al 2 O 3 (s) First, let’s convert g FeO to g Fe: We can produce 19.4 g Fe if FeO is limiting g FeO × 3 mol Fe 3 mol FeO 1 mol FeO g FeO × 1 mol Fe g Fe × = 19.4 g Fe Mass Limiting Reactant Problem

34 Chapter FeO(l) + 2 Al(l) → 3 Fe(l) + Al 2 O 3 (s) Second, lets convert g Al to g Fe: We can produce 77.6 g Fe if Al is limiting g Al × 3 mol Fe 2 mol Al 1 mol Al g Al × 1 mol Fe g Fe × = 77.6 g Fe Problem, continued

35 Chapter Let’s compare the two reactants: –25.0 g FeO can produce 19.4 g Fe –25.0 g Al can produce 77.6 g Fe FeO is the limiting reactant. Al is the excess reactant. Problem Finished

36 Chapter Limiting reactant problems involving volumes follow the same procedure as those involving masses, except we use volumes. volume reactant  volume product We can convert between the volume of the reactant and the product using the balanced equation. Volume Limiting Reactant Problems

37 Chapter How many liters of NO 2 gas can be produced from 5.00 L NO gas and 5.00 L O 2 gas? 2 NO(g) + O 2 (g) → 2 NO 2 (g) Convert L NO to L NO 2, and L O 2 to L NO 2 : = 5.00 L NO L NO × 2 L NO 2 2 L NO = 10.0 L NO L O 2 × 2 L NO 2 1 L O 2 Volume Limiting Reactant Problem

38 Chapter Let’s compare the two reactants: –5.00 L NO can produce 5.00 L NO 2 –5.00 L O 2 can produce 10.0 L NO 2 NO is the limiting reactant. O 2 is the excess reactant. Problem, continued

39 Chapter When you perform a laboratory experiment, the amount of product collected or physically weighed by a balance is the actual yield. The amount of product calculated from a limiting reactant problem or a balanced chemical equation is the theoretical yield. The percent yield is the amount of the actual yield compared to the theoretical yield. × 100 % = percent yield actual yield theoretical yield **Percent Yield**

40 Chapter Suppose a student performs a reaction and obtains g of CuCO 3 and the theoretical yield is g. What is the percent yield? Cu(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) → CuCO 3 (s) + 2 NaNO 3 (aq) The percent yield obtained is 88.6%. × 100 % = 88.6 % g CuCO g CuCO 3 Calculating Percent Yield

41 Chapter The coefficients in a balanced chemical reaction are the mole ratio of the reactants and products. The coefficients in a balanced chemical reaction are the volume ratio of gaseous reactants and products. We can convert moles or liters of a given substance to moles or liters of an unknown substance in a chemical reaction using the balanced equation. Chapter Summary

42 Chapter Here is a flow chart (i.e. road map) for doing stoichiometry problems. *****Chapter Summary, continued*****

43 Chapter The limiting reactant is the reactant that is used up first in a chemical reaction. The theoretical yield of a reaction is the amount calculated based on the limiting reactant. The actual yield is the amount of product isolated in an actual experiment. The percent yield is the ratio of the actual yield to the theoretical yield. Chapter Summary, continued


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