Section 9.1 Using Chemical Equations (Stoichiometry) Unit 2: Chemical reactions Mole Ratios & Stochiometry Section 9.1 Using Chemical Equations (Stoichiometry) Objectives 1. To understand the information given in a balanced equation 2. To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Section 9.1 Using Chemical Equations (Stoichiometry) Quinoa Side Dish Section 9.1 Using Chemical Equations (Stoichiometry) 1. Information Given by Chemical Equations • The coefficients of a balanced equation give the relative numbers of molecules. • A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction. Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • A balanced equation can predict the moles of product that a given number of moles of reactants will yield. How many moles 02 from __ moles H2O? Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation. Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • • • • • The mole ratio is determined by: Mole ratio = new mole substance A balanced mole substance A How many mole ___ from ___ moles ___? 2H2O 2H2 + O2 • mole mole ratio mole Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • Must first have a BALANCED equation • __C3H8(g) + __O2 __CO2(g) + __H2O(g) + heat • Calculate the number of moles of CO2 formed when 4.30 mol C3H8 combust. • mole mole ratio mole Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • From this new mole value, we can calculate the new recipe! • C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat • 4.30 C3H8 +__O212.9 CO2 + __ H2O + heat • Calculate the number of moles of H2O formed when 4.30 mol C3H8 combust. • mole mole ratio mole Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • For all reactants AND products! • C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat • 4.30 C3H8 +__O212.9 CO2 + 17.2 H2O + heat • Calculate the number of moles of O2 consumed when 4.30 mol C3H8 combust. • mole mole ratio mole Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • C3H8(g) + 5O2 3CO2(g) + 4H2O(g) + heat • Calculate the number of moles of all reactants and products when 2.70 mol C3H8 combust. • 2.7 C3H8 +__O2 __ CO2 + __ H2O + heat • 2.7 C3H8 +13.5 O2 8.1 CO2 + 10.8 H2O + heat Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • Balance the following equation and determine how much NH3 can be made from 1.3 mol H2. • __N2 + __H2 __NH3 • N2 + 3H2 2NH3 • How much N2 will be needed to completely react the 1.3 mol H2? Section 9.1 Using Chemical Equations (Stoichiometry) 2. Mole-mole Relationships • Using the balanced equation, determine the new mole recipe given the new mole value. • 2 K + 2 H2O H2 + 2 KOH • ___K + 1.7 H2O ___ H2 + ___KOH • 1.7 K + 1.7 H2O 0.85 H2 + 1.7 KOH • 4 NH3 + 5 O2 4 NO + 6 H2O • ___ NH3 + ___ O2 0.6 NO + ___ H2O • 0.6 NH3 + 0.75 O2 0.6 NO + 0.9 H2O Section 9.1 Using Chemical Equations (Stoichiometry) Objectives Review 1. To understand the information given in a balanced equation 2. To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Section 9.2 Using Chemical Equations to Calculate Mass Objectives 1. To learn to use Stoichiometry to relate masses of reactants and products in a chemical reaction 2. To perform mass calculations that involve scientific notation 3. Calculate the theoretical predications for a reaction that will be performed in the lab Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • Stoichiometry is the process of using a balanced chemical equation to determine the relative masses of reactants and products involved in a reaction. – Scientific notation can be used for the masses of any substance in a chemical equation. • Mole ratio = new mole substance A • balanced mole substance A Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • N2 + 3H2 2NH3 • How much NH3 will be produced when 2.6 g H2 completely reacts with N2? • **Can’t do a mass ratio!!** • Must first convert from grams to moles. • Don’t forget the mole! • grams • mole grams mole ratio mole Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • N2 + 3H2 2NH3 • How much NH3 will be produced when 2.6 g H2 completely reacts with N2? • 2.6 grams H2 • mole grams mole ratio • 0.867 mol NH3: 14.71 g NH3 mole Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • N2 + 3H2 2NH3 • __ N2 + 1.3 H2 0.867 NH3 (new mol recipe) • __ g N2 2.6 g H2 14.71 g NH3 (new g recipe) • How much N2 will be consumed when 2.6 g H2 completely reacts? • Use the same mole ratio— • 0.433 mol N2: 12.12 g N2 Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • N2 + 3H2 2NH3 • 0.433 N2 + 1.3 H2 0.867 NH3 (new mol recipe) • 12.12 g N2 2.6 g H2 14.71 g NH3 (new g recipe) • How does this relate to the Law of Conservation of Mass? • Does the mass reactants = mass products? • 12.12 g + 2.6 g = 14.72 g • Compared to 14.71 g Why the 0.01 g difference? Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations Section 9.2 Using Chemical Equations to Calculate Mass 1. Mass Calculations • __Al(S) + __I2(g) __AlI3(s) • Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed. • grams grams • mole mole ratio mole (next) Section 9.2 Using Chemical Equations to Calculate Mass 2. Scientific Notation • • • • • • __LiOH(s) + __CO2(g) __Li2CO3(s) + __H2O(l) What mass of CO2 can 1 X 103 g LiOH absorb? What mass of H2O will be available for use? What mass of Li2CO3 will be generated? First Balance! Then, below!! grams grams • mole mole ratio mole Section 9.2 Using Chemical Equations to Calculate Mass 2. Scientific Notation (CO2 scrubber on space vehicles) • • • • • • • • • 2 LiOH(s) + CO2(g) Li2CO3(s) + H2O(l) 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O 1 X 103 g 915.2 g 1539.2 g 374.4 g What mass of CO2 can 1 X 103 g LiOH absorb? What mass of H2O will be available for use? What are some possible uses for the H2O? What mass of Li2CO3 will be generated? What will be done with the Li2CO3? Conserved? Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations Using Scientific Notation • Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with silica, SiO2, in the glass to produce gaseous silicon tetrafluoride and liquid water. The unbalanced reaction is: • __HF(aq) + __SiO2(s) __SiF4(g) + __H2O(l) • Calculate the mass of HF that is needed and the products that are produced to react completely with 5.68 g SiO2. Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations Comparisons • Which is the better antacid? Baking Soda,NaHCO3, Mg(OH)2? • How many moles of HCl will react with 1.00 g of each antacid? • NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) Mg(OH)2(s) + 2HCl(aq) 2H2O(l) + MgCl2(aq) • grams • mole grams mole ratio mole Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations for Lab- don’t round too much! • Predict the charged balanced products and their solubility. Then balance the equation, and calculate the number of grams of all reactants and products needed to completely react with 0.01 mol SrCl2. • ___SrCl2 + ___Na2CO3 • Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations for lab • SrCl2 + Na2CO3 SrCO3 + 2NaCl • 0.01 SrCl2 + ___Na2CO3___SrCO3+___NaCl • 0.01SrCl2 + 0.01Na2CO3 0.01SrCO3 + 0.02NaCl • 1.59 g + 1.06 g • Conservation of mass? • Theoretical vs Real… 1.48 g + 1.17 g Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations for lab • Calculate the amount of grams contained in 0.01 mol the following hydrated crystals: • SrCl2 x 6H2O • Na2CO3 X H2O 3 0 Section 9.2 Using Chemical Equations to Calculate Mass 2. Mass Calculations for lab • 0.01SrCl2 x 6H2O + 0.01Na2CO3 X H2O 0.01SrCO3 + 0.02NaCl • 2.67 g + 1.24 g 1.48 g + 1.17 g Section 9.2 Using Chemical Equations to Calculate Mass Objectives Review 1. To understand the information given in a balanced equation 2. To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) 3. Calculate the theoretical predications for a reaction that will be performed in the lab Section 9.3 Limiting Reactants and Percent Yield Objectives 1. To understand the concept of limiting reactants 2. To learn to recognize the limiting reactant in a reaction 3. To learn to use the limiting reactant to do stoichiometric calculations 4. To learn to calculate percent yield Section 9.3 Limiting Reactants and Percent Yield 1. The Concept of Limiting Reactants • Stoichiometric mixture (balanced) – N2(g) + 3H2(g) 2NH3(g) Section 9.3 Limiting Reactants and Percent Yield 1. The Concept of Limiting Reactants • Limiting reactant mixture (runs out first) – N2(g) + 3H2(g) 2NH3(g) Section 9.3 Limiting Reactants and Percent Yield Mixture of CH4 and H2O Molecules Reacting CH4 + H2O 3H2 + CO Section 9.3 Limiting Reactants and Percent Yield CH4 + H2O 3H2 + CO Section 9.3 Limiting Reactants and Percent Yield Limiting Reactants • • • • The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Limiting and Excess Reactants depend on the actual amounts of reactants present and must be calculated for each different reaction. (Water won’t always be in excess!) Section 9.3 Limiting Reactants and Percent Yield Limiting Reactants Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant • What mass of water is needed to react with 249 g methane (CH4)? • CH4 + H2O 3H2 + CO • 249 g CH4 • How many g of H2 and CO are produced? • 279 g H2O, 93 g H2, 434 g CO Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant • How many g of H2 and CO are produced when 249 g methane (CH4) reacts with 300 g H2O? • CH4 + H2O 3H2 + CO • 249 g + 279 g 93 g + 434 g (last slide) • 249 g + 300 g 93 g + 434 g • The 249 g CH4 will react with only 279 g H2O, thereby leaving 21 g H2O in excess. CH4 is the reactant that will run out first and is the LIMITING REACTANT. Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant • So, how do we figure this out using Stoich? • 1) Balance Reaction • 2) Do Stoich to relate EACH reactant to one product • 3) Whichever reactant produces the LEAST amount of product is the LIMITING REACTANT Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant • Suppose 2.50 X 104 g N2 reacts with 5 X 103 g H2. Determine the limiting reactant. • ___N2 + ___H2 ___NH3 • 2.50 X 104 g N2 • 5 X 103 g H2 • From N2 1785.72 mol NH3 • From H2 1666.67 mol NH3 ; 28333.3 g NH3 • Not just because original mass of H2 was less. Section 9.3 Limiting Reactants and Percent Yield 2. Calculations Involving a Limiting Reactant • How much N2 will be produced when 18.1 g NH3 and 90.4 g CuO react? Determine the limiting reactant. • __NH3 + __CuO __N2 + __Cu + __H2O • 18.1 g NH3 • 90.4 g CuO • NH3 0.53 mol N2 • CuO 0.38 mol N2 *limiting reactant* 10.64 g N2 Section 9.3 Limiting Reactants and Percent Yield 3. Percent Yield • Theoretical Yield – The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. Just what we’ve been calculating! • The actual yield (amount produced is usually given) of a reaction is usually less than the max theoretical yield because of side or incomplete reactions. • Percent Yield The actual amount of a given product as the percentage of the theoretical yield. Section 9.3 Limiting Reactants and Percent Yield 3. Percent Yield • From the previous set of calculations, we determined that 10.64 g of N2 would theoretically be produced from the reaction: • 2NH3 + 3CuO N2 + 3Cu + 3H2O • What would the Percent Yield be if 7.25 g of N2 were actually produced? • • 7.25 g N2 10.64 g N2 X 100 = 68.14 % Section 9.3 Limiting Reactants and Percent Yield 3. Percent Yield • Consider the following reaction: • __CO(g) + __H2(g) __CH3OH(l) • If 6.85 X 104 g CO reacts with 8.6 X 103 g H2 to produce an actual yield of 3.57 X 104 g CH3OH, determine the limiting reactant, the theoretical yield, and the percent yield. Section 9.3 Limiting Reactants and Percent Yield 3. Percent Yield • CO(g) + • 2446.4 mol CO • 2446.4 mol CH3OH 2H2(g) CH3OH(l) 4300 mol H2 2150 mol CH3OH *Limiting • 2150 mol CH3OH *32g/mol = 68800 g CH3OH • % yield = 3.57 X 104 g/ 6.88 X 104 g = 51.89 % • 28 g/mol CO: 2 g/mol H2: 32 g/mol CH3OH Section 9.3 Limiting Reactants and Percent Yield 3. Percent Yield • Consider the following reaction: • __TiCl4 + __O2 __TiO2 + __Cl2 • If 6.71 X 103 g TiCl4 reacts with 2.45 X 103 g O2 with a percent yield of 75%, determine the limiting reactant, the theoretical yield, and the actual yield of TiO2. • TiCl4 is limiting • 2119.6 g TiO2 is actually produced in this reaction. • 189.68 g/mol TiCl4: 32 g/mol o2: 79.88 g/mol TiO2 Section 9.3 Limiting Reactants and Percent Yield Objectives Review 1. To understand the concept of limiting reactants 2. To learn to recognize the limiting reactant in a reaction 3. To learn to use the limiting reactant to do stoichiometric calculations 4. To learn to calculate percent yield 5. Work Session: Page 308 2, 3, 4, 5