Chapter 6 Chemical Reactions and Quantities Percent Yield and Limiting Reactants

Vocabulary Theoretical Yield Actual Yield Percent Yield Limiting Reactants Excess

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

Calculating Percent Yield Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns, and you have to throw them out. The rest of the cookies you make 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 With a limited amount of 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 of CO are produced when 30.0 g of O2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

Solution 3) 76.2% yield STEP 1 Given: 40.0 g of CO produced (actual) 30.0 g of O2 used Need: percent yield of CO STEP 2 Write a plan to calculate % yield of CO: g of O2 moles of moles of g of CO O2 CO (theoretical) Percent yield of CO = g of CO (actual) x 100% g of CO (theoretical)

Solution (continued) STEP 3 Write conversion factors: 1 mole of O2 = 32.0 g of O2 1 mole O2 and 32.0 g O2 32.0 g O2 1 mole O2 1 mole of O2 = 2 moles of CO 1 mole O2 and 2 moles CO 2 moles CO 1 mole O2 1 mole of CO = 28.0 g of CO 1 mole CO and 28.0 g CO 28.0 g CO 1 mole CO

Solution (continued) STEP 4 Setup to calculate theoretical yield in g of O2: 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 of CO (theoretical) Setup to calculate percent yield: 40.0 g CO (actual) x 100 = 76.2% yield (3) 52.5 g CO (theoretical)

Learning Check When N2 and 5.00 g of H2 are mixed, the reaction produces 16.0 g of NH3. What is the percent yield for the reaction? N2(g) + 3H2(g) 2NH3(g) 1) 31.3% of NH3 2) 56.9% of NH3 3) 80.0% of NH3

Solution 2) 56.9% STEP 1 Given: 16.0 g of NH3 produced (actual) 2) 56.9% STEP 1 Given: 16.0 g of NH3 produced (actual) 5.00 g of H2 used Need: percent yield of NH3 STEP 2 Write a plan to calculate % yield of NH3: g of H2 moles of moles of g of NH3 H2 NH3 (theoretical) Percent yield of NH3 = g of NH3 (actual) x 100% g of NH3 (theoretical)

Solution (continued) STEP 3 Write conversion factors: 1 mole of H2 = 2.02 g of H2 1 mole H2 and 2.02 g H2 2.02 g H2 1 mole H2 1 mole of H2 = 2 moles of NH3 1 mole H2 and 2 moles NH3 2 moles NH3 1 mole H2 1 mole of NH3 = 17.0 g of NH3 1 mole NH3 and 17.0 g NH3 17.0 g NH3 1 mole NH3

Solution (continued) STEP 4 Setup to calculate theoretical yield of g of NH3: 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 = 28.1 g of NH3 (theoretical) Setup to calculate percent yield: Percent yield = 16.0 g NH3 x 100 = 56.9% yield (2) 28.1 g NH3

Limiting Reactant A limiting reactant in a chemical reaction is the substance that is used up 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?

Reacting Amounts (continued) 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.

Example 1 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 2 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

Limiting Reactant When 4.00 moles of H2 is mixed with 2.00 moles of Cl2, how many moles of HCl can form? H2(g) + Cl2(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 smaller number of moles of product.

Limiting Reactant (continued) HCl from H2 4.00 moles H2 x 2 moles HCl = 8.00 moles of HCl 1 moles H2 HCl from Cl2 2.00 moles Cl2 x 2 moles HCl = 4.00 moles of HCl 1 mole Cl2 4.00 moles of HCl is the smaller number of moles produced. Thus, Cl2 will be used up. The limiting reactant is Cl2.

Check Calculations Equation Initially H2 4.00 moles Cl2 2.00 moles 2HCl 0 mole Reacted/ Formed –2.00 moles +4.00 moles Left after reaction (4.00 – 2.00) Excess 0 moles (2.00 – 2.00) Limiting (0 + 4.00) Product possible

Limiting Reactants 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 of Ca3N2 3 moles Ca (Ca is used up) Moles of Ca3N2 from N2 2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles of Ca3N2 1 mole N2 Ca is used up. Thus, Ca is the limiting reactant.

Learning Check What is the mass of water that can be produced when 8.00 g of H2 and 24.0 g of O2 react? 2H2(g) + O2(g) 2H2O(l) 1) 8.0 g of H2O 2) 27.0 g of H2O 3) 72 g of H2O

Solution Moles of H2O from H2: 3) 72 g of H2O Moles of H2O from H2: 8.00 g H2 x 1 mole H2 x 2 moles H2O = 4.0 moles of H2O 2.0 g H2 2 moles H2 Moles of H2O from O2: 24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles of H2O 32.0 g O2 1 mole O2 Smaller number of moles of H2O 1.50 moles H2O x 18.0 g H2O = 27.0 g of H2O 1 mole H2O