Chapter 5 Chemical Reactions Law of Conservation of Mass and Quantities 5.7 Mole Relationships in Chemical Equations

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1 Chapter 5 Chemical Reactions and Quantities Law of Conservation of Mass 5.7 Mole Relationships in Chemical Equations The Law of Conservation of Mass indicates that in an ordinary chemical reaction, Matter cannot be created or destroyed. No change in total mass occurs in a reaction. Mass of products is equal to mass of reactants. 1 Conservation of Mass 2 Reading Equations with Moles Consider the following equation: Fe(s) + 3O2(g) 2Fe2O3(s) This equation can be read in moles by placing the word moles between each coefficient and formula. + Copyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings Reactants 2 moles Ag + 1 moles S 2 (107.9 g) + 1(32.1 g) 27.9 = = Products 1 mole Ag2S 1 (27.9 g) = 27.9 g moles Fe + 3 Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for any two Consider the following equation: 3 H2(g) + N2(g) 2 NH3(g) substances in an equation. Fe(s) + Fe and O2 Fe and Fe2O3 O2 and Fe2O3 3O2(g) 2Fe2O3(s) moles Fe and moles Fe and and A. A mole-mole factor for H2 and N2 is 1) 3 moles N2 2) 1 mole N2 1 mole H2 3 moles H2 moles Fe moles Fe 3) 1 mole N2 2 moles H2 B. A mole-mole factor for NH3 and H2 is 1) 1 mole H2 2) 2 moles NH3 3) 3 moles N2 2 moles NH3 3 moles H2 2 moles NH

2 Calculations with Mole Factors How many moles of Fe 2 O 3 can be produced from 6.0 moles O 2? Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship: 3 mole O 2 = 2 mole Fe 2 O 3 Write a mole-mole factor to determine the moles of Fe 2 O 3. How many moles of Fe are needed for the reaction of 12.0 moles O 2? Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe 6.0 mole O 2 x 2 mole Fe 2 O 3 =.0 moles Fe 2 O 3 3 mole O Chapter 5 Chemical Reactions and Quantities 5.8 Mass Calculations for Reactions Moles to Grams Suppose we want to determine the mass (g) of NH 3 that can form from 2.50 moles N 2. N 2 (g) + 3H 2 (g) 2NH 3 (g) The plan needed would be moles N 2 moles NH 3 grams NH 3 The factors needed would be: mole factor NH 3 /N 2 and the molar mass NH Moles to Grams The setup for the solution would be: 2.50 mole N 2 x 2 moles NH 3 x 17.0 g NH 3 1 mole N 2 1 mole NH 3 given mole-mole factor molar mass = 85.0 g NH 3 How many grams of O 2 are needed to produce 0.00 mole Fe 2 O 3 in the following reaction? Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 1) 38. g O 2 2) 19.2 g O 2 3) 1.90 g O

3 Calculating the Mass of a Reactant Calculating the Mass of a Reactant The reaction between H 2 and O 2 produces 13.1 g water. How many grams of O 2 reacted? 2H 2 (g) + O 2 (g) 2H 2 O(g)? g 13.1 g The plan and factors would be g H 2 O mole H 2 O mole O 2 g O 2 mass H 2 O factor mass O 2 The setup would be: 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O g H 2 O 2 moles H 2 O 1 mole O 2 = 11.6 g O 2 mass H 2 O factor mass O Calculating the Mass of Product Acetylene gas C 2 H 2 burns in the oxyacetylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g CO 2? When 18.6 g ethane gas C 2 H 6 burns in oxygen, how many grams of CO 2 are produced? 2C 2 H 2 (g) + 5O 2 (g) 1) 88.6 g C 2 H 2 2).3 g C 2 H 2 3) 22.2 g C 2 H 2 CO 2 (g) + 2H 2 O(g) 2C 2 H 6 (g) + 7O 2 (g) 18.6 g? g The plan and factors would be CO 2 (g) + 6H 2 O(g) g C 2 H 6 mole C 2 H 6 mole CO 2 g CO 2 mass C 2 H 6 factor mass CO Calculating the Mass of Product The setup would be 18.6 g C 2 H 6 x 1 mole C 2 H 6 x moles CO 2 x.0 g CO g C 2 H 6 2 moles C 2 H 6 1 mole CO 2 mass C 2 H 6 factor mass CO 2 How many grams H 2 O are produced when 35.8 g C 3 H 8 react by the following equation? C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + H 2 O(g) g H 2 O g H 2 O g H 2 O = 5. g CO

4 Chapter 5 Chemical Reactions and Quantities 5.9 Percent Yield and Limiting Reactants Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product, which is calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g) Calculating Percent Yield You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burn and you throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 8 cookies to eat Percent yield 8 cookies x 100 = 80% yield 60 cookies 21 Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O 2 (g) 2CO(g) What is the percent yield if 0.0 g CO are produced when 30.0 g O 2 are used? 1) 25.0% 2) 75.0% 3) 76.2% 22 Limiting Reactant When N 2 and 5.00 g H 2 are mixed, the reaction produces 16.0 g NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 1) 31.3 % 2) 56.5 % 3) 80.0 % 2NH 3 (g) A limiting reactant in a chemical reaction is the substance that Is used up first. Limits the amount of product that can form and stops the reaction. 23 2

5 Reacting Amounts Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. Only place settings are possible. Initially Plates 5 Forks 6 Spoons Knives 7 How many table settings are possible from 5 plates, 6 forks, spoons, and 7 knives? Used Left over What is the limiting item? The limiting item is the spoon. 25 Example of An Everyday Limiting Reactant 26 Example of An Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. With 8 slices of bread, only sandwiches could be made. The bread is the limiting item Limiting Reactant Limiting Reactant When.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form? H2(g) + Cl2(g) 2HCl (g) HCl from H2.00 moles 2.00 moles.00 moles H2 x 2 moles HCl = 8.00 moles HCl 1 moles H2 (not possible) HCl from Cl moles Cl2 x 2 moles HCl =.00 moles HCl 1 mole Cl2 (smaller number of moles, Cl2 will be used up first)??? moles Calculate the moles of product that each reactant, H2 and Cl2, could produce. The limiting reactant is the one that produces the smallest amount of product. The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl

6 Check Calculations Limiting Reactants Using Mass Initially H 2 Reacted/ Formed Left after reaction.00 moles Cl moles 2HCl 0 mole moles moles +.00 moles 2.00 moles Excess 0 mole Limiting.00 moles If.80 moles Ca are mixed with 2.00 moles N 2, which is The limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) moles of Ca 3 N 2 from Ca.80 moles Ca x 1 mole Ca 3 N 2 = 1.60 moles Ca 3 N 2 3 moles Ca (Ca is used up) moles of Ca 3 N 2 from N moles N 2 x 1 mole Ca 3 N 2 = 2.00 moles Ca 3 N 2 1 mole N 2 (not possible) Ca is used up when 1.60 mole Ca 3 N 2 forms. Thus, Ca is the limiting reactant Limiting Reactants Using Mass Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 2.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) Moles H 2 O from H 2 : 8.00 g H 2 x 1 mole H 2 x 2 moles H 2 O = 3.97 moles H 2 O 2.02 g H 2 2 moles H 2 (not possible) Moles H 2 O from O 2 : 2.0 g O 2 x 1 mole O 2 x 2 moles H 2 O = 1.50 moles H 2 O 32.0 g O 2 1 mole O 2 O 2 is limiting The maximum amount of product is 1.50 moles H 2 O, which is converted to grams moles H 2 O x 18.0 g H 2 O = 27.0 g H 2 O 1 mole H 2 O 3 6