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Chapter 8 Quantities in Chemical Reactions. Products are carbon dioxide and w a t e r. Octane in gas tank. Octane mixes with oxygen. 2006, Prentice Hall. CHAPTER OUTLINE. Global Warming: Too Much Carbon Dioxide.
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Chapter 8 Quantities in Chemical Reactions Products are carbon dioxide and water Octane in gas tank Octane mixes with oxygen 2006, Prentice Hall
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 CO2. • Consider the combustion of octane (C8H18), a component of gasoline: 2 C8H18(l)+ 25 O2(g) 16 CO2(g)+ 18 H2O(g) • The balanced chemical equation shows that 16 mol of CO2 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 CO2 levels have risen 25%
The Source of Increased CO2 • the primary source of the increased CO2 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
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.
MOLARRATIOS For example: 1 N2 (g) + 3 H2 (g) 2 NH3 (g) This is the molar ratios between the reactants and products 1 molecule 3 molecules 2 molecules 100 molecules 300 molecules 200 molecules 106 molecules 3x106 molecules 2x106 molecules 1 mole 3 moles 2 moles
Examples: Determine each mole ratio below based on the reaction shown: 2 C4H10 + 13 O2 8 CO2 + 10 H2O
Stoichiometry • Stoichiometry - Allows us to predict products that form in a reaction based on amount of reactants. 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. 2H2 + O2 2H2O
STOICHIOMETRICCALCULATIONS Mass-mass calculations • Stoichiometric calculations can be classified as one of the following: Mass-mole calculations Mole-mole calculations MASS of compound B MASS of compound A MM MM MOLES of compound A molar ratio MOLES of compound B
MOLE-MOLECALCULATIONS • Relates moles of reactants and products in a balanced chemical equation MOLES of compound A molar ratio MOLES of compound B
Example 1: How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N2 present) 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 2 32 mol H2 = 21 mol NH3 3 Mole ratio
Example 2: In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 3 6.80 mol NH3 = 10.2 mol H2 2 Mole ratio
MASS-MOLECALCULATIONS • Relates moles and mass of reactants or products in a balanced chemical equation MASS of compound A MM MOLES of compound A molar ratio MOLES of compound B
Example 1: How many moles of ammonia can be produced from the reaction of 125 g of nitrogen? 1 N2 (g) + 3 H2 (g) 2 NH3 (g) 1 2 125 g N2 = 8.93 mol NH3 1 28.0 Molar mass Mole ratio
MASS -MASSCALCULATIONS • Relates mass of reactants and products in a balanced chemical equation MASS of compound B MASS of compound A MM MM MOLES of compound A molar ratio MOLES of compound B
Example 1: What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown? 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) Moles of carbon dioxide Mass of propane Moles of propane Mass of carbon dioxide
1 3 44.1 1 44.0 1 Example 1: 1 C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 175 g C3H8 = 524 g CO2 Molar mass Mole ratio Molar mass
LIMITINGREACTANT • 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.
LIMITING REACTANT ANALOGY Consider the following recipe for a sundae:
Limiting reactant Excess reactants 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.
LIMITINGREACTANT • When solving limiting reactant problems, assume each reactant is limiting reactant, and calculate the desired quantity based on that assumption. • Compare your answers for each assumption; the lower value is the correct assumption. Lower value is correct A + B C Calculate amount of C A is LR Calculate amount of C B is LR
Limiting reactant Mass-mass calculations Example 1: A fuel mixture used in the early days of rocketry was a mixture of N2H4 and N2O4, as shown below. How many grams of N2 gas is produced when 100 g of N2H4 and 200 g of N2O4 are mixed? 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)
Example 1: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2H4 is LR 1 3 100 g N2H4 2 32.04 4.68 mol N2
Example 1: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2O4 is LR 1 3 200 g N2O4 1 92.00 6.52 mol N2
Example 1: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Assume N2H4 is LR 4.68 mol N2 Assume N2O4 is LR 6.52 mol N2 N2H4 is LR Correct amount
Example 1: 2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g) Calculate mass of N2 28.0 4.68 mol N2 = 131 g N2 1
Limiting Reactant Example 2: How many grams of AgBr can be produced when 50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as shown below: MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2
Example 2: MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2 Assume MgBr2 is LR 1 2 50.0 g MgBr2 184.1 1 187.8 102 g AgBr 1
Example 2: MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2 Assume AgNO3 is LR 1 2 100.0 g AgNO3 169.9 2 187.8 111 g AgBr 1
Example 2: MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2 Assume MgBr2 is LR 102 g AgBr Assume AgNO3 is LR 111 g AgBr MgBr2 is LR Correct amount
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.
PERCENT YIELD • The percent yield of a reaction is obtained as follows:
Example 1: 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? 92.7 %
Actual yield Example 2: Silicon carbide can be formed from the reaction of sand (SiO2) with carbon as shown below: 1 SiO2 (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?
Example 2: 1 SiO2 (s) + 3 C (s) 1 SiC (s) + 2 CO (g) Calculate theoretical yield 1 1 40.1 100 g SiO2 60.1 1 1 66.7 g SiC
Example 2: Calculate percent yield 77.1 %
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.
Chap. 8 terms you should know • 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. actual yield % yield = x 100 theoretical yield