Percent Yield and Limiting Reactants Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings Edited by bbg

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)

Guide to Calculations for Percent Yield Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies

Learning Check Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O2(g) 2CO(g) What is the percent yield if 40.0 g CO are produced when 30.0 g O2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

Solution 3) 76.2 % yield theoretical yield of CO 30.0 g O2 x 1 mole O2 x 2 moles CO x 28.0 g CO 32.0 g O2 1 mole O2 1 mole CO = 52.5 g CO (theoretical) percent yield 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

Learning Check When N2 and 5.00 g H2 are mixed, the reaction produces 16.0 g NH3. What is the percent yield for the reaction? N2(g) + 3H2(g) 2NH3(g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

Solution 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3 2.02 g H2 3 moles H2 1 mole NH3 = 28.2 g NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.7 % 28.2 g NH3

Limiting Reactant 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.

Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Reacting Amounts Only 4 place settings are possible. Initially Used Left over Plates 5 4 1 Forks 6 4 2 Spoons 4 4 0 Knives 7 4 3 The limiting item is the spoon. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

Example of An Everyday Limiting Reactant 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.

Guide to Calculating Product from a Limiting Reactant Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Limiting Reactant When 4.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form? H2(g) + Cl(g)  2HCl (g) 4.00 moles 2.00 moles ??? 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.

Limiting Reactant HCl from H2 4.00 moles H2 x 2 moles HCl = 8.00 moles HCl 1 moles H2 (not possible) HCl from Cl2 2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl 1 mole Cl2 (smaller number of moles, Cl2 will be used up first) The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

Check Calculations Initially H2 4.00 moles Cl2 2.00 moles 2HCl 0 mole Reacted/ Formed -2.00 moles +4.00 moles Left after reaction Excess 0 mole Limiting

Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N2, which is the limiting reactant? 3Ca(s) + N2(g)  Ca3N2(s)moles of Ca3N2 from Ca 4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2 3 moles Ca (Ca is used up) moles of Ca3N2 from N2 2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2 1 mole N2 (not possible) Ca is used up when 1.60 mole Ca3N2 forms. Thus, Ca is the limiting reactant.

Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H2 and 24.0 g O2 react? 2H2(g) + O2(g) 2H2O(l)

Limiting Reactants Using Mass Moles H2O from H2: 8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O 2.02 g H2 2 moles H2 (not possible) Moles H2O from O2: 24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O 32.0 g O2 1 mole O2 O2 is limiting The maximum amount of product is 1.50 moles H2O, which is converted to grams. 1.50 moles H2O x 18.0 g H2O = 27.0 g H2O 1 mole H2O