9.1 Interpreting a chemical Equation CHAPTER 9: STOICHIOMETRY H 2 (g) + Cl 2 (g) 2 HCl (g) 1 molecule 1 molecule 2 molecules N 2 + 3 H 2 (g) 2 NH 3 (g) molecule(s) molecule(s) molecule(s) It follows that any multiples of these coefficients will be in same ratio! 2 H 2 (g) + O 2 (g) 2 H 2 O (g) X 1000 molecule(s) molecule(s) molecule(s) X N molecule(s) molecule(s) molecule(s) Since N=Avogadro s # = 6.02x10 23 molecules = 1 mol 2 H 2 (g) + O 2 (g) 2 H 2 O (g) mol (s) mol (s) mol (s) Thus, the coefficients in a chemical equation give the mole ratios of reactants and products in a reaction. Give the mole ratios for each of the following: 1. H 2 (g) + Cl 2 (g) 2 HCl (g) mol(s) mol(s) mol(s) 2. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) mol(s) mol(s) mol(s) mol(s) 1
9.2 MOLE-MOLE RELATIONSHIPS Consider the following reaction: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) 1. Use unit factors to determine how many moles of O 2 are needed to completely react with 2.25 moles of C 3 H 8. 2. How many moles of CO 2 form when 3.50 moles of O 2 react? 3. How many moles of H 2 O form when 4.75 moles of CO 2 form? 4. How many moles of C 3 H 8 are required to produce 1.50 moles of H 2 O? 2
9.3 Stoichiometry Problems stoichiometry (STOY-key-OM-etry): -refers to the amounts of reactants and products in a chemical reaction -a stoichiometry problem generally involves relating amounts of reactants and/or products to each other in terms of moles 9.4 Mass-Mass (Stoichiometry) Problems MASS OF KNOWN Molar Mass MOLES OF KNOWN Mole-Mole Ratio MOLES OF UNKNOWN Molar Mass MASS OF UNKNOWN Example: Consider the mass of CO needed to react completely with 50.0 g of Fe 2 O 3. Fe 2 O 3 (s) + 3 CO 2 Fe (s) + 3 CO 2 (g) 1. Calculate the mass of CO needed to react completely with 50.0 g of Fe 2 O 3. 2. Calculate the mass of iron produced when 123 g of CO reacts completely. 3. Calculate the mass of CO2 produced when 75.0 g of iron is produced. 3
9.5 The Limiting Reactant Concept (LIMITING REAGENT) In practice, reactants will not always be present in the exact amounts necessary for all reactants to be converted completely into products. Some reactants (usually the least expensive) are present in larger amounts and are never completely used up reactant(s) in excess Only in a limited supply of the other reactants (usually the more expensive) are present, so these are completely used up limiting reactant since it limits the amount of product that can be made MAKING BICYCLES -Parts needed: - 1 bicycle frame - 1 seat - 2 pedals - 2 wheels (rims + tires) Example: How many bicycles can be made with 5 frames, 6 seats, 15 pedals and 8 wheels? (Indicate the limiting reactant and reactants in excess.) 4
2 cups Bisquick MAKING BISQUICK TM PANCAKES 1 cup milk 14 pancakes 2 eggs 2 cups of Bisquick TM + 1 Cup milk + 2 eggs 14 pancakes Example: If you have 10 cups of Bisquick TM, 10 cups of milk, and 12 eggs, how many pancakes can you make? (Indicate the limiting reagent(s) and reagent(s) in excess.) 9.6 Limiting Reactant Problems GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS: 1. Calculate the amount (moles or mass) of product formed using the amount of each reactant given -Use mass-mass conversions Smallest amount = amount of product formed! 2. Whichever reactant produces the smaller amount of product limiting reagent 3. All other reactant(s) in excess 5
Ex. 1: Consider the reaction between aluminum metal and hydrochloric acid to produce hydrogen gas: 2 Al (s) + 6 HCl (aq) 2 AlCl 3 (aq) + 3 H 2 (g) Calculate the number of moles of hydrogen gas produced when 5.00 moles of aluminum metal react with 5.00 moles of HCl. Limiting reactant = Reactant in excess = Ex. 2: Consider the reaction fro propane (C 3 H 8 ) burning: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Calculate the number of moles of Carbon dioxide gas produced. When 1.50 moles of propane react with 5.00 moles of oxygen. Limiting reactant = Reactant in excess = 6
actual yield 9.7 Percent yield = x 100% theoretical yield theoretical yield: Amount of product one should get based on the chemical equation and the amount of reactants present -One generally calculates this in grams from info given Actual yield: Amount of produce one actually obtains -Generally smaller than the theoretical yield because of impurities and other adverse conditions in the lab -This value has to be provided in a problem Ex.: Calculate the percent yield for a reaction with a theoretical yield of 75.0 g of carbon dioxide if the actual amount of carbon dioxide produced was 59.2 g. 7
Example: Consider the following equation: 2 K + Cl 2 2 KCl a. How many grams of KCl is produced from 2.50 g of K and excess Cl 2. b. From 1.00 g of Cl 2 and excess K? c. When 2.50 g of K and 1.00 g Cl 2 react together, the mass of KCl produced is, the limiting reactant is, and the reactant in excess is. d. Calculate the percent yield if 86.7 g of KCl is actually produced when 2.50 g of K and 1.00g Cl 2 react. 8
9.6 Interpreting a chemical Equation CHAPTER 9: STOICHIOMETRY H 2 (g) + Cl 2 (g) 2 HCl (g) 1 molecule 1 molecule 2 molecules N 2 + 3 H 2 (g) 2 NH 3 (g) _1 molecule(s) 3 molecule(s) 2 molecule(s) It follows that any multiples of these coefficients will be in same ratio! 2 H 2 (g) + O 2 (g) 2 H 2 O (g) X 1000 _2000 molecule(s) _1000 molecule(s) _2000 molecule(s) X N 1.204 x 10 23 molecule(s) 6.02 x 10 23 molecule(s) 1.204 x 10 23 molecule(s) Since N=Avogadro s # = 6.02x10 23 molecules = 1 mol 2 H 2 (g) + O 2 (g) 2 H 2 O (g) _2 mol (s) 1 mol (s) 2 mol (s) Thus, the coefficients in a chemical equation give the mole ratios of reactants and products in a reaction. Give the mole ratios for each of the following: 1. H 2 (g) + Cl 2 (g) 2 HCl (g) 1_mol(s) 1_mol 2_mol(s) 2. C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) _1 mol(s) 5_mol(s) _3 mol(s) 4_mol(s) 9
9.7 MOLE-MOLE RELATIONSHIPS Consider the following reaction: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) 4. Use unit factors to determine how many moles of O 2 are needed to completely react with 2.25 moles of C 3 H 8. 2.25 mol C 3 H 8 x 5 mol O 2 = 11.3 mol O 2 1 mol C 3 H 8 5. How many moles of CO 2 form when 3.50 moles of O 2 react? 3.50 mol O 2 x 3 mol CO 2 = 2.10 mol CO 2 5 mol O 2 6. How many moles of H 2 O form when 4.75 moles of CO 2 form? 4.74 mol CO 2 x 4 mol H 2 O = 6.33 mol H2O 3 mol CO 2 4. How many moles of C 3 H 8 are required to produce 1.50 moles of H 2 O? 1.50 mol H 2 O x 1 mol C 3 H 8 = 0.375 mol C 3 H 8 4 mol H 2 O 10
9.8 Stoichiometry Problems stoichiometry (STOY-key-OM-etry): -refers to the amounts of reactants and products in a chemical reaction -a stoichiometry problem generally involves relating amounts of reactants and/or products to each other in terms of moles 9.9 Mass-Mass (Stoichiometry) Problems MASS OF KNOWN Molar Mass MOLES OF KNOWN Mole-Mole Ratio MOLES OF UNKNOWN Molar Mass MASS OF UNKNOWN Example: Consider the mass of CO needed to react completely with 50.0 g of Fe 2 O 3. Fe 2 O 3 (s) + 3 CO 2 Fe (s) + 3 CO 2 (g) 4. Calculate the mass of CO needed to react completely with 50.0 g of Fe 2 O 3. Given: 50.0 g Fe 2 O 3 Find: g CO Format: g Fe 2 O 3 mol Fe 2 O 3 mol CO g CO 50.0 g Fe 2 O 3 x 1 mol Fe 2 O 3 x 3 mol CO x 28.0 g CO = 26.3 g CO 160. g Fe 2 O 3 1 mol Fe 2 O 3 1 mol CO 5. Calculate the mass of iron produced when 123 g of CO reacts completely. Given: 123 g CO Find: g Fe Format: g CO mol CO mol Fe g Fe 123 g CO x 1 mol CO x 2 mol Fe x 55.8 g Fe = 163 g Fe 28.0 g CO 3 mol CO 1 mol Fe 6. Calculate the mass of CO 2 produced when 75.0 g of iron is produced. Given: 75.0 g Fe Find: g CO 2 Format: g Fe mol Fe mol CO 2 g CO 2 75.0 g Fe x 1 mol Fe x 3 mol CO 2 x 44.0 g CO 2 = 88.7 g CO 2 55.8 g Fe 2 mol Fe 1 mol CO 2 11
9.10 The Limiting Reactant Concept (LIMITING REAGENT) In practice, reactants will not always be present in the exact amounts necessary for all reactants to be converted completely into products. Some reactants (usually the least expensive) are present in larger amounts and are never completely used up reactant(s) in excess Only in a limited supply of the other reactants (usually the more expensive) are present, so these are completely used up limiting reactant since it limits the amount of product that can be made MAKING BICYCLES -Parts needed: - 1 bicycle frame - 1 seat - 2 pedals - 2 wheels (rims + tires) Example: How many bicycles can be made with 5 frames, 6 seats, 15 pedals and 8 wheels? (Indicate the limiting reactant and reactants in excess.) 1 bicycle frame + 1 seat + 2 pedals + 2 wheels = 1 bicycle 5 frames x 1 bike = 5 bikes 1 frame 6 seats x 1 bike = 6 bikes 1 seat 15 pedals x 1 bike = 7.5 bikes 2 pedals 12
8 wheels x 1 bike = 4 bikes 2 wheels Limiting reactant wheels Excess reactants Bicycle frames, seats, pedals 2 cups Bisquick MAKING BISQUICK TM PANCAKES 1 cup milk 14 pancakes 2 eggs 2 cups of Bisquick TM + 1 Cup milk + 2 eggs 14 pancakes Example: If you have 10 cups of Bisquick TM, 10 cups of milk, and 12 eggs, how many pancakes can you make? (Indicate the limiting reagent(s) and reagent(s) in excess.) 2 c Bisquick + 1 C milk + 2 eggs = 14 pancakes 10 c. Bisquick x 14 pancakes = 70 pancakes 2 c. Bisquick 10 c milk x 14 pancakes = 140 pancakes 1 c milk 12 eggs x 14 pancakes = 84 pancakes 2 eggs Limiting reagent = Bisquick Excess reagent = milk, eggs 13
9.6 Limiting Reactant Problems GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS: 4. Calculate the amount (moles or mass) of product formed using the amount of each reactant given -Use mass-mass conversions Smallest amount = amount of product formed! 5. Whichever reactant produces the smaller amount of product limiting reagent 6. All other reactant(s) in excess Ex. 1: Consider the reaction between aluminum metal and hydrochloric acid to produce hydrogen gas: 2 Al (s) + 6 HCl (aq) 2 AlCl 3 (aq) + 3 H 2 (g) Calculate the number of moles of hydrogen gas produced when 5.00 moles of aluminum metal react with 5.00 moles of HCl. 2 Al + 6 HCl 2 AlCl 3 + 3 H 2 5.00 mol Al x 3 mol H 2 = 7.5 mol H 2 2 mol Al 5.00 mol HCl x 3 mol H 2 = 2.5 mol H 2 6 mol HCl Limiting reactant = HCl Reactant in excess = Al Ex. 2: Consider the reaction fro propane (C 3 H 8 ) burning: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) Calculate the number of moles of Carbon dioxide gas produced. When 1.50 moles of propane react with 5.00 moles of oxygen. C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O 1.50 mol C 3 H 8 x 3 mol CO 2 = 4.5 mol CO 2 1 mol C 3 H 8 14
5.00 mol O 2 x 3 mol CO 2 = 3 mol CO 2 5 mol O 2 Limiting reactant = O 2 Reactant in excess = C 3 H 8 actual yield 9.7 Percent yield = x 100% theoretical yield theoretical yield: Amount of product one should get based on the chemical equation and the amount of reactants present -One generally calculates this in grams from info given Actual yield: Amount of produce one actually obtains -Generally smaller than the theoretical yield because of impurities and other adverse conditions in the lab -This value has to be provided in a problem Ex.: Calculate the percent yield for a reaction with a theoretical yield of 75.0 g of carbon dioxide if the actual amount of carbon dioxide produced was 59.2 g. 59.2 g CO 2 x 100 = 78.9% 75.0 g CO 2 15
Example: Consider the following equation: 2 K + Cl 2 2 KCl a. How many grams of KCl is produced from 2.50 g of K and excess Cl 2. Given: 2.50 g K Find: g KCl Format: g K mol K mol KCl g KCl 2.50 g K x 1 mol K x 2 mol KCl x 74.6 g KCl = 4.77 g KCl 39.1 g K 2 mol K 1 mol KCl b. From 1.00 g of Cl 2 and excess K? Given: 1.00 g Cl 2 Find: g KCl Format: g Cl 2 mol Cl 2 mol KCl g KCl 1.00 g Cl 2 x 1 mol K x 2 mol KCl x 74.6 g KCl = 2.10 g KCl 70.9 g Cl 2 1 mol Cl 2 1 mol KCl c. When 2.50 g of K and 1.00 g Cl 2 react together, the mass of KCl produced is 2.10 g, the limiting reactant is Cl 2, and the reactant in excess is K. d. Calculate the percent yield if 86.7 g of KCl is actually produced when 2.50 g of K and 1.00g Cl 2 react. 16
86.7 x 100 = 4130% 2.10 17