Chapter 8 Quantities in Chemical Reactions 2006, Prentice Hall Octane in gas tank Octane mixes with oxygen Products are carbon dioxide and water

2 CHAPTER OUTLINE  Stoichiometry  Molar Ratios  Mole-Mole Calculations  Mass-Mole Calculations  Mass-Mass Calculations  Limiting Reactant  Percent Yield

Global Warming: Too Much Carbon Dioxide The combustion of fossil fuels such as octane (shown here) produces water and carbon dioxide as products. Carbon dioxide is a greenhouse gas that is believed to be responsible for global warming.

The greenhouse effect Greenhouse gases act like glass in a greenhouse, allowing visible-light energy to enter the atmosphere but preventing heat energy from escaping. Outgoing heat is trapped by greenhouse gases.

Combustion of fossil fuels produces CO 2. Consider the combustion of octane (C 8 H 18 ), a component of gasoline: 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) The balanced chemical equation shows that 16 mol of CO 2 are produced for every 2 mol of octane burned.

Global Warming scientists have measured an average 0.6°C rise in atmospheric temperature since 1860 during the same period atmospheric CO 2 levels have risen 25%

The Source of Increased CO 2 the primary source of the increased CO 2 levels are combustion reactions of fossil fuels we use to get energy (methane and octane) –1860 corresponds to the beginning of the Industrial Revolution in the US and Europe

8 STOICHIOMETRY  Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation.  A balanced chemical equation provides several important information about the reactants and products in a chemical reaction.

9 MOLAR RATIOS For example: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 1 molecule3 molecules2 molecules 100 molecules300 molecules200 molecules 10 6 molecules3x10 6 molecules2x10 6 molecules 1 mole3 moles2 moles This is the molar ratios between the reactants and products

10 Examples: Determine each mole ratio below based on the reaction shown: 2 C 4 H O 2  8 CO H 2 O

Stoichiometry - Allows us to predict products that form in a reaction based on amount of reactants. Stoichiometry The amount of each element must be the same throughout the overall reaction. For example, the amount of element H or O on the reactant side must equal the amount of element H or O on the product side. 2H 2 + O 2 2H 2 O

12 STOICHIOMETRIC CALCULATIONS  Stoichiometric calculations can be classified as one of the following: MOLES of compound A MOLES of compound B molar ratio Mole-mole calculations MASS of compound A MM Mass-mole calculations MASS of compound B MM Mass-mass calculations

13 MOLE-MOLE CALCULATIONS  Relates moles of reactants and products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio

14 How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N 2 present) Example 1: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 32 mol H = 21 mol NH 3 Mole ratio

15 In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment? Example 2: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 6.80 mol NH = 10.2 mol H 2 Mole ratio

16 MASS-MOLE CALCULATIONS  Relates moles and mass of reactants or products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio MASS of compound A MM

17 How many moles of ammonia can be produced from the reaction of 125 g of nitrogen? Example 1: 1 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) 125 g N = 8.93 mol NH Molar mass Mole ratio

18 MASS -MASS CALCULATIONS  Relates mass of reactants and products in a balanced chemical equation MOLES of compound A MOLES of compound B molar ratio MASS of compound A MM MASS of compound B MM

19 What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown? Example 1: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) Mass of propane Moles of propane Moles of carbon dioxide Mass of carbon dioxide

20 Example 1: 1 C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) 175 g C 3 H = 524 g CO Molar mass Mole ratio Molar mass

21 LIMITING REACTANT  When 2 or more reactants are combined in non- stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess.  This reactant is referred to as limiting reactant.  When doing stoichiometric problems of this type, the limiting reactant must be determined first before proceeding with the calculations.

22 LIMITING REACTANT ANALOGY Consider the following recipe for a sundae:

23 LIMITING REACTANT ANALOGY How many sundaes can be prepared from the following ingredients: The number of sundaes possible is limited by the amount of syrup, the limiting reactant. Limiting reactant Excess reactants

24 LIMITING REACTANT  When solving limiting reactant problems, assume each reactant is limiting reactant, and calculate the desired quantity based on that assumption. A + B  C A is LR B is LR Calculate amount of C  Compare your answers for each assumption; the lower value is the correct assumption. Lower value is correct

25 A fuel mixture used in the early days of rocketry was a mixture of N 2 H 4 and N 2 O 4, as shown below. How many grams of N 2 gas is produced when 100 g of N 2 H 4 and 200 g of N 2 O 4 are mixed? Example 1: 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Limiting reactant Mass-mass calculations

26 Example 1: 100 g N 2 H mol N N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 H 4 is LR 1

27 Example 1: 200 g N 2 O mol N N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 O 4 is LR 1

28 Example 1: 6.52 mol N 2 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Assume N 2 O 4 is LR Assume N 2 H 4 is LR 4.68 mol N 2 Correct amount N 2 H 4 is LR

29 Example 1: 4.68 mol N = 131 g N 2 2 N 2 H 4 (l) + 1 N 2 O 4 (l)  3 N 2 (g) + 4 H 2 O (g) Calculate mass of N 2 1

30 How many grams of AgBr can be produced when 50.0 g of MgBr 2 is mixed with g of AgNO 3, as shown below: Example 2: MgBr AgNO 3  2 AgBr + Mg(NO 3 ) 2 Limiting Reactant

31 Example 2: 50.0 g MgBr Assume MgBr 2 is LR 1 MgBr AgNO 3  2 AgBr + Mg(NO 3 ) g AgBr

32 Example 2: g AgNO Assume AgNO 3 is LR 1 MgBr AgNO 3  2 AgBr + Mg(NO 3 ) g AgBr

33 Example 2: 111 g AgBr Assume AgNO 3 is LR Assume MgBr 2 is LR 102 g AgBr Correct amount MgBr 2 is LR MgBr AgNO 3  2 AgBr + Mg(NO 3 ) 2

34 PERCENT YIELD  The amount of product calculated through stoichiometric ratios are the maximum amount product that can be produced during the reaction, and is thus called theoretical yield.  The actual yield of a product in a chemical reaction is the actual amount obtained from the reaction.

35 PERCENT YIELD  The percent yield of a reaction is obtained as follows:

36 In an experiment forming ethanol, the theoretical yield is 50.0 g and the actual yield is 46.8 g. What is the percent yield for this reaction? Example 1: 92.7 %

37 Silicon carbide can be formed from the reaction of sand (SiO 2 ) with carbon as shown below: Example 2: 1 SiO 2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g) When 100 g of sand are processed, 51.4g of SiC is produced. What is the percent yield of SiC in this reaction? Actual yield

38 Example 2: 100 g SiO g SiC 1 1 Calculate theoretical yield 1 1 SiO 2 (s) + 3 C (s)  1 SiC (s) + 2 CO (g)

39 Example 2: 77.1 % Calculate percent yield

Theoretical and Actual Yield In order to determine the theoretical yield, we use reaction stoichiometry to determine the amount of product each of our reactants could make. The theoretical yield will always be the least possible amount of product. –The theoretical yield will always come from the limiting reactant. Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield.

1.Limiting reactant - the reactant that limits the amount of product produced in a chemical reaction. The reactant that makes the least amount of product. 2.Theoretical yield - the amount of product that can be made in a chemical reaction based on the amount of limiting reactant. 3.Actual yield - the amount of product actually produced by a chemical reaction. 4.Percent yield - The percent of the theoretical yield that was actually obtained. % yield = actual yield theoretical yield x 100 Chap. 8 terms you should know

42 THE END