Chapter 10. Mole-Mole Stoichiometry. Example 10/21/2011. Chemistry Relationships in Chemical Reactions

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1 10/1/011 Chapter 10 Chemistry Relationships in Chemical Reactions Mole-Mole Stoichiometry When writing a balanced chemical equation, one can make relationships between any two substances in the reaction and perform quantitative analysis (moles, grams, etc.). This is known as STOICHIOMETRY. Let s look at an example!!! Example K (s) + HCl (aq) KCl (aq) + H (g) One can make relationships between any two substances. In this equation, there are 1 relationships one can observe (all in units of moles): Given Wanted Wanted Given = Wanted Given 1

2 10/1/011 Example K (s) + HCl (aq) KCl (aq) + H (g) One can make relationships between any two substances. In this equation, there are many relationships one can observe (all in units of moles): moles K moles HCl moles HCl moles KCl moles K moles KCl moles HCl 1mole H moles K 1mole H moles KCl moles K moles HCl moles K moles KCl moles HCl Example K (s) + HCl (aq) KCl (aq) + H (g) One can make relationships between any two substances. In this equation, there are 1 relationships one can observe (all in units of moles): moles KCl 1mole H 1mole H moles K 1mole H moles HCl 1mole H moles KCl Example Mole-Mole Stoichiometry K (s) + HCl (aq) KCl (aq) + H (g) How many moles of hydrogen are produced from. moles of HCl? Key Conv. Factor: 1mole H moles HCl Solution: 1mole H. moles HCl = moles HCl 1.61mole H

3 10/1/011 Grams-Grams Stoichiometry In solving gram-gram calculations, one must know the molecular mass of the given substance and wanted substance in addition to the mole-to-mole ratio between substances. Let s look at the general setup!!! Figure 10.-Conversion among grams, moles and number of particles for two species in a chemical change. MM is molar mass and N A = number of molecules or atoms (as particles). Gram-Gram Stoichiometry To perform this type of stoichiometry, one must remember these skills from earlier chapters: 1. You will write chemical formulas.. You will calculate molar masses from chemical formulas.. You will use molar masses to change mass to moles and vice versa. 4. You will write and balance chemical equations. 5. You will use the equation to change from moles of one species to moles of another.

4 10/1/011 Gram-Gram Stoichiometry Mass-Mass Path: Mass of Given of Given of Wanted Mass of Wanted mass Given mol Given mass Given Step1 mol Wanted mass Wanted mol Given mol Wanted Step Step Example Gram-Gram Stoichiometry Determine how much carbon dioxide will be produced from 4.96g of propane. REACTION: C H 8 (g) + 5O (g) CO (g) + 4H O (g) Solution REACTION: C H 8 (g) + 5O (g) CO (g) + 4H O (g) Given: Propane is C H 8 Initial Mass of Propane = 4.96g Molar Mass (C H 8 ) = 44.11g/mol Mole Ratio: 1 mole C H 8 = mole CO Molar Mass (CO ) = 44.01g/mol Need: Mass of CO (in grams) 4

5 10/1/011 Solution (cont d) REACTION: C H 8 (g) + 5O (g) CO (g) + 4H O (g) Solution setup: g C H 8 mol C H 8 mol CO g CO 1mol C H mol CO 44.01g CO 4.96g C 8 H 8 =? 44.11g C H 8 1mol C H 8 1mol CO Mass (CO )=14.8g Percent Yield In this case, percent yield (or percent efficiency) is a ratio between the actual yield (in grams) and the theoretical yield (in grams) as shown in the formula below: Actual Yield Theoretical Yield 100% = Percent Yield ILLUSTRATION Percent Yield Figure 10.. A barium chloride solution is added to a solution of a sodium sulfate solution. (a) Separation of the clear, colorless solutions of sodium sulfate and barium chloride before mixing. (b) the barium chloride solution is added to the sodium sulfate solution. A white precipitate of barium sulfate forms. (c) Barium sulfate is separated from the solution by filtration. (d) The actual yield of barium sulfate is determined by drying the product and subtracting the mass of the filter paper from the combined mass of the filter paper and product. 5

6 10/1/011 Percent Yield (cont d) When the actual yield is less the theoretical yield of the reaction, here are some reason for this difference: 1. reactants may be impure. the reaction may not go to completion. other side reactions may occur Practice Problem REACTION: Mg (s) + O (g) MgO (s) I have 5.6 grams of magnesium which result in obtaining 51. grams of magnesium oxide. What is the percent yield of this reaction? Solution REACTION: Mg (s) + O (g) MgO (s) Given: Need: Magnesium is Mg Initial Mass of Magnesium = 5.6g Molar Mass (Mg) = 4.1g/mol Mole Ratio: moles Mg = moles MgO Molar Mass (MgO) = 40.1g/mol Actual Yield = 51.g MgO a) Mass of MgO(in grams)-theoretical Yield b) % Yield of this reaction 6

7 10/1/011 Solution (cont d) REACTION: Mg (s) + O (g) MgO (s) Solution setup: g Mg mol Mg mol MgO g MgO 1mol Mg mol MgO 40.1g MgO 5.6g Mg =? 4.1g Mg mol Mg 1mol MgO Mass (MgO)=59.0g Solution (cont d) REACTION: Mg (s) + O (g) MgO (s) Solution setup: Percent Yield Actual (MgO)=51.g Theoretical (MgO)=59.0g 51.g 100% = 86.9% Yield 59.0g Percent Yield A Conversion Factor In the example, the percent yield was 86.9% for MgO. The percent yield can be expressed as a conversion factor. 86.9g MgO (actual) 100g MgO (theoretical) or 100g MgO (theoretical) 86.9g MgO (actual) 7

8 10/1/011 Problem #5 The Haber process for making ammonia from nitrogen in the air is given by the equation N + H NH. Calculate the mass of hydrogen that must be supplied to make 5.00 x 10 kg of ammonia in a system that has an 88.8% yield. KEY: 88.8g NH (actual) 100g NH (theoretical) ANSWER: 100. kg H or 100g NH (theoretical) 88.8g NH (actual) ILLUSTRATION Limiting Reagent (or Reactant) C O CO (START) 0 (USED) (AFTER) 1 0 C + O CO C O CO (START) 0 (USED) (AFTER) 1 0 This illustrates the following: moles of C when reacts with moles of O leads to moles of CO produced. C has an extra mole unused where all of oxygen is consumed. What term do we call for this behavior in a chemical system. 8

9 10/1/011 C + O CO (cont d) C O CO (START) 0 (USED) (AFTER) 1 0 Oxygen is classified as the limiting reagent because all of this substance is consumed according to this chemical equation. C + O CO (cont d) C O CO (START) 0 (USED) (AFTER) 1 0 Carbon is deemed the excess reactant because there is some amount of unreacted substance remaining after the completion of the reaction. Procedure Limiting Reactant Problem (Grams-Grams) 1. Convert the number of grams of each reactant to moles.. Identify the limiting reactant.. Calculate the number of moles of each species that reacts or is produced. 4. Calculate the number of moles of each species that remains after the reaction. 5. Change the number of moles of each species to grams. 9

10 10/1/011 Practice Problem In a chemical system, the following reaction is observed: AgNO (aq) + K CO (aq) Ag CO (s) + KNO (aq) If 5.44 moles of silver nitrate mixes with.44 moles of potassium carbonate, what amounts (in grams) of each substances will remain and formed after the reaction. AgNO (aq) + K CO (aq) Ag CO (s) + KNO (aq) AgNO K CO Ag CO KNO (START) (USED) (AFTER) Identify the limiting reactant: moles AgNO.44 moles K CO = 1mole K CO 6.88 moles AgNO 1mole K CO 5.44 moles AgNO = moles AgNO.7 moles K CO AgNO (aq) + K CO (aq) Ag CO (s) + KNO (aq) AgNO K CO Ag CO KNO (START) (USED) (AFTER) Since AgNO is the limiting reactant, one would expect that all of its composition will be consumed while K CO will be the excess reactant. 10

11 10/1/011 Additional Problem In a chemical system, the following reaction is observed: CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl (aq) If 1.g of calcium chloride reacts with 5.99g of sodium phosphate, determine (1) what will be the limiting reactant and () amounts (in grams) for each substance after completion of the reaction. CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl (aq) CaCl Na PO 4 Ca (PO 4 ) 6NaCl Grams (START) (START) (USED)???? (AFTER) Identify the limiting reactant: 1mole CaCl 1.g CaCl = moles CaCl g CaCl 1mole Na PO 5.99g Na PO 4 4 = moles Na PO 16.94g Na PO 4 4 CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl (aq) CaCl Na PO 4 Ca (PO 4 ) 6NaCl Grams (START) (START) (USED)???? (AFTER) Identify the limiting reactant (cont d): Na PO 4 is the limiting reagent Convert moles of CaCl to Na PO 4. moles Na PO moles CaCl 4 = 0.07 moles Na PO moles CaCl 4 1mole Na PO 5.99g Na PO 4 4 = moles Na PO g Na PO 4 11

12 10/1/011 CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl (aq) CaCl Na PO 4 Ca (PO 4 ) 6NaCl Grams (START) (START) (USED) (AFTER) moles CaCl moles Na PO4 = moles CaCl moles Na PO 4 6 moles NaCl moles Na PO4 = moles NaCl moles Na PO 4 1mole Ca (PO ) moles Na PO4 4 = 0.018moles Ca (PO4) moles Na PO 4 CaCl (aq) + Na PO 4 (aq) Ca (PO 4 ) (s) + 6NaCl (aq) CaCl Na PO 4 Ca (PO 4 ) 6NaCl Grams (START) (START) (USED) (AFTER) g CaCl moles CaCl = 5.0g CaCl 1mole CaCl 58.44g NaCl moles NaCl = 6.4g NaCl 1 mole NaCl 10.18g Ca (PO ) 0.018moles Ca (PO ) 4 = 1mole Ca (PO ) g Ca (PO4) AgNO (aq) + K CO (aq) Ag CO (s) + KNO (aq) AgNO K CO Ag CO KNO (START) (USED) (AFTER) Since AgNO is the limiting reactant, one would expect that all of its composition will be consumed while K CO will be the excess reactant. 1

13 10/1/011 Limiting Reactants Smaller Amount Method One can also use stoichiometry to determine which reactant leads to the smaller amount of product based on the initial amount of the given substance. Let s look at this in detail!!! Procedure Limiting Reactant Problem (Lowest Amount) 1. Calculate the amount of product that can be formed by the initial amount of each reactant. a) The reactant that yields the smaller amount of product is the limiting reactant. b) The smaller amount of product is the amount that will be formed when all of the limiting reactant is used up.. Calculate the amount of excess reactant that is used by the total amount of limiting reactant.. Subtract from the amount of the excess reactant present initially the amount that is used by all of the limiting reactant. The difference is the amount of excess reactant that is left. Practice Problem 5.44g of solid aluminum reactants with.98g of aqueous copper(ii) nitrate yields an unknown amount of copper and aqueous aluminum nitrate. What mass of aluminum, copper(ii) nitrate and copper are observed once the reaction has completed? 1

14 10/1/011 Step 1: Balance the chemical equation. Solution Reaction (UNBALANCED): Al (s) + Cu(NO ) (aq) Cu (s) + Al(NO ) (aq) REACTANTS PRODUCTS 1 Al, 1 Cu, NO 1 Al, 1 Cu, NO Multiply nitrate compounds by and respectively. Al (s) + Cu(NO ) (aq) Cu (s) + Al(NO ) (aq) REACTANTS PRODUCTS 1 Al, Cu, 6NO Al, 1 Cu, 6NO Multiply Al by and Cu by. Al (s) + Cu(NO ) (aq) Cu (s) + Al(NO ) (aq) REACTANTS PRODUCTS Al, Cu, 6NO Al, Cu, 6NO Solution (cont d) Reaction: Al (s) + Cu(NO ) (aq) Cu (s) + Al(NO ) (aq) Step : Find the mass of solid copper (in grams). From aluminum: 1mole Al moles Cu 6.55g Cu 5.44g Al = 19. g Cu 6.98g Al moles Al 1 mol Cu From copper(ii) nitrate: 1mole Cu(NO ) moles Cu 6.55g Cu.98g Cu(NO ) g Cu(NO ) moles Cu(NO ) 1 mol Cu Limiting Reactant: Cu(NO Final Amount: ) 1.01g Cu = 1.01g Cu Solution (cont d) Reaction: Al (s) + Cu(NO ) (aq) Cu (s) + Al(NO ) (aq) Step : Determine the amount of excess reactant remaining. From aluminum: 1mole Cu moles Al 6.98g Al 1.01g Cu = 0.86g 6.55g Cu moles Cu 1mol Al Al: 5.44g initial -0.86g used = 5.15g remaining From copper(ii) nitrate: Since all of this compound has been consumed, there will be 0g. Final Amount: 1.01g Cu 14

15 10/1/011 Energy (J) Energy is defined as the capacity to perform work. The SI unit for energy is joule (J). Jouleis an amount of energy exerted when a force of one newton (force required to cause a mass of 1 kg to accelerate at a rate of 1 m/s ) is applied over a displacement of one meter: 1 joule = 1 newton 1 meter 1 J = 1 kg m /sec Energy (cal) Other units of energy are in calories (cal) and kilocalories (kcal). Calorie is defined as the amount of energy required to raise the temperature of one gram of water per 1 o C. 1 cal joules 4.184J/cal Also, the following relationship is also true: 1 kcal kj 4.184kJ/kcal Energy (cal) (cont d) Additionally, there is a unit of energy for nutrition called Calorie (Cal). Typically, a diet for a normal individual is about 000 Calories (kcal) per day. 1 Cal I kcal 1000 cal 1 Cal kj 4184 J An individual who consumes 000 Cal is equilvalent to over 800 kj/diet. 15

16 10/1/011 Sample Question Determine the number of kilocalories, calories, and joules in 54.1 kj of energy. joules: 1000J kJ = J 1kJ calories: kilocalories: 4 1cal J = cal 4.184J 4 1kcal cal = 1.9kcal 1000cal Thermochemical Equations Enthalpy has a representative symbol known as H. Heat of reaction refers to the amount of heat which is either given off or absorbed in a chemical reaction usually measured in laboratory. A heat of reaction, commonly known as the enthalpy of reaction ( H) refers to a change in energy based on if the system gives off heat or absorbs heat for a system. Thermochemical Equations (cont d) If energy is released from the system to the surroundings, it is classified as an exothermic reaction. EXAMPLE: C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) H = -855 kj or C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) kj 16

17 10/1/011 Thermochemical Equations (cont d) If energy is absorbed from the surroundings to the system, it is classified as an endothermic reaction. EXAMPLE: NH (g) + 9kJ N (g) + H (g) H = +9 kj or NH (g) + 9kJ N (g) + H (g) Practice Problem Write the two thermochemical reactions for the process when dinitrogen tetroxide, N O 4 (g) converts to nitrogen dioxide, NO (g), an endothermic reaction requiring 54 kj per moles of dinitrogen tetroxide. Thermochemical Stoichiometry Figure 10.5-Conversion among grams or energy and H, moles, and the number of particles for two species in a chemical change. 17

18 10/1/011 Sample Problem Problem #64 How many grams of hydrogen would have to react to produce 71.9 kj of energy from the reaction H (g) + O (g) H O (g) kj? Key Conversion Factor: 484 kj = mol H Solution H (g) + O (g) H O (g) kj mol H.016 g 71.9kJ = 0.599g H 484 kj 1mol H 18