Chapter 9 Chemical Quantities in Reactions. Mole Relationships in Chemical Equations Copyright ©...

Chapter 9

Chemical Quantities in Reactions

Mole Relationships in Chemical Equations

Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

• The equation can be read in “moles” by placing the word “______” or “______” after each coefficient.

4Fe(s) + 3O2(g) 2Fe2O3(s)

4 mol Fe + 3 mol O2 2 mol Fe2O3

Moles in Equations

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)_____ molecules NH3 react with _____ molecules O2

to produce

_____ molecules NO and ____ molecules H2O

and

_____ mol NH3 react with ______ mol O2

to produce

______ mol NO and ______ mol H2O

Quantities in A Chemical Reaction

• Take another look at this equation

4Fe(s) + 3O2(g) 2Fe2O3(s)

4 mol Fe + 3 mol O2 2 mol Fe2O3

Moles in Equations

A mole-mole factor is a ______ of the moles for two

substances in an equation.

4Fe(s) + 3O2(g) 2Fe2O3(s)

Fe and O2 4 mol Fe and 3 mol O2

3 mol O2 4 mol Fe

Fe and Fe2O3 4 mol Fe and 2 mol Fe2O3

2 mol Fe2O3 4 mol Fe

O2 and Fe2O3 3 mol O2 and 2 mol Fe2O3

2 mol Fe2O3 3 mol O2

Writing Mole-Mole Factors

Consider the following equation:

3H2(g) + N2(g) 2NH3(g)

1. A mole factor for H2 and N2 is

2. A mole factor for NH3 and H2 is

Learning Check

How many moles of Fe2O3 can form from 6.0 mol O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

STEP 1 Given 6.0 mol O2 Need: moles of Fe2O3.

STEP 2 moles O2 moles Fe2O3

STEP 3 3 mol O2 = 2 mol Fe2O3

3 mol O2 and 2 mol Fe2O3


STEP 4 Set up problem using the mol factor.

6.0 mol O2 x 2 mol Fe2O3 = 4.0 mol Fe2O3

3 mol O2

Calculations with Mole Factors

How many moles of Fe are needed to react with

12.0 mol O2?

4Fe(s) + 3O2(g) 2 Fe2O3(s)

12.0 mol O2 x 4 mol Fe = 3 mol O2

Learning Check

16.0 mol of Fe

Conservation of Mass

The Law of Conservation of Mass indicates

• No change in _____ _____ occurs in a reaction.

• Mass of products is ______ to mass of reactants.

Mass Calculations for Reactions


Stoichiometry

• Chemical Stoichiometry: using mass and quantity relationships among reactants and products in a chemical reaction to make predictions about how much product will be made.

In an ordinary chemical reaction,

• Matter cannot be created nor destroyed.

• The number of atoms of each element are equal.

• The mass of reactants equals the mass of products.

H2(g) + Cl2(g) 2HCl(g)

2 mol H, 2 mol Cl = 2 mol HCl atoms atoms molecules 2(1.008) + 2(35.45) = 2(36.46) 72.92 g = 72.92 g

Law of Conservation of Mass

Conservation of Mass2 Ag + S → Ag2S

2 mol Ag + 1 mol S = 1 mol Ag2S

2 (107.9 g) + 1(32.07 g) = 1 (247.9 g)

247.9 g reactants = 247.9 g product


Stoichiometric Mantra

Grams A to Moles A to Moles B to Grams B

1

2

3

The reaction between H2 and O2 produces 13.1 g H2O.

How many grams of O2 reacted?

2H2(g) + O2(g) 2H2O(g)

STEP 1 Given 13.1 g H2O Need grams O2

STEP 2 Plan: g H2O mol H2O mol O2 g O2

STEP 3 1 mol O2 = 2 mol H2O 1 mol H2O = 18.02 g

1 mol O2 = 32.00 g O2

STEP 4 13.1 g H2O x 1 mol H2O x 1 mol O2 x 32.00 g O2

18.02 g H2O 2 mol H2O 1 mol O2

=

Calculating the Mass of a Reactant

11.6 g O2

How many grams of O2 are needed to produce 0.400 mol Fe2O3?

4Fe(s) + 3O2(g) 2 Fe2O3(s)

mole factor molar mass0.400 mol Fe2O3 x 3 mol O2 x 32.00 g O2


= 19.2 g O2

Learning Check

Learning CheckAcetylene gas C2H2 burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g CO2?

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

C 12.01 g/mol

H 1.008 g/mol

O 16.00 g/mol

Limiting Reactants


Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that

• Is used up first.

• Stops the reaction.

• Limits the amount of product that can form.

Reacting Amounts

In a table setting, there is 1

plate, 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

Four table settings can be made.

Initially Use 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 of Everyday Limiting Reactant

How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter?

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

Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

Example of 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.

Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

Limiting ReactantsWhen 4.00 mol H2 is mixed with 2.00 mol

Cl2,how

many moles of HCl can form?

H2(g) + Cl2(g) 2HCl (g)

4.00 mol 2.00 mol ??? mol

• Calculate the moles of product from each reactant, H2 and Cl2.

• The limiting reactant is the one that produces the smaller amount of product.

Limiting Reactants Using Moles H2(g) + Cl2(g) 2HCl (g)

4.00 mol 2.00 mol ??? mol

HCl from H2

4.00 mol H2 x 2 mol HCl = 8.00 mol HCl 1 mol H2

HCl from Cl2

2.00 mol Cl2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl2

The limiting reactant is Cl2 because it produces the smaller number of moles of HCl and is used up first.

(not possible)

(smaller amount)

Checking Calculations

Initially H2

4.00 mol

Cl2 2.00 mol

2HCl

0 mol

Reacted/

Formed

-2.00 mol -2.00 mol +4.00 mol

Left after reaction

2.00 mol

Excess

0 mol

Limiting

4.00 mol

Limiting Reactants Using Mass

If 4.80 mol Ca mixed with 2.00 mol N2, which is thelimiting reactant? 3Ca(s) + N2(g) Ca3N2(s)

Moles of Ca3N from Ca

4.80 mol Ca x 1 mol Ca3N2 = 1.60 mol Ca3N2 3 mol Ca

Moles of Ca3N2 from N2

2.00 mol N2 x 1 mol Ca3N2 = 2.00 mol Ca3N2

1 mol N2

All Ca is used up when 1.60 mol Ca3N2 forms. Thus, Ca is the limiting reactant. N2 is in excess.

(smaller amount)

(not possible)

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)

What do we need to do?

Determine amount of water is produced for each of the reactants.

Limiting Reactants Using MassCalculate the mass of water produced when 8.00 g H2 and

24.0 g O2 react?

2H2(g) + O2(g) 2H2O(l)

Calculate the grams of H2O produced for each reactant.

H2: 8.00 g H2 x 1 mol H2 x 2 mol H2O x 18.02 g H2O

2.016 g H2 2 mol H2 1 mol H2O

= 71.5 g H2O

O2:

24.0 g O2 x 1 mol O2 x 2 mol H2O x 18.02 g H2O 32.00 g O2 1 mol O2 1 mol H2O

= 27.0 g H2O

O2 is the limiting reactant.

(not possible)

(smaller amount)

Limiting Reactants Using MassCalculate the masses of reactants when (1) 8.00 g H2 and (2)

24.0 g O2 react?

2H2(g) + O2(g) 2H2O(l)Calculate the grams of other reactant needed for each reactant.

H2: 8.00 g H2 x 1 mol H2 x 1 mol O2 x 32.00 g O2

2.016 g H2 2 mol H2 1 mol O2

= 63.5 g O2

O2:

24.0 g O2 x 1 mol O2 x 2 mol H2 x 2.016 g H2 32.00 g O2 1 mol O2 1 mol H2

= 3.02 g H2

O2 is the limiting reactant.

Some H2 remains in excess.

(not possible)

(is possible)

Percent Yield


Theoretical, Actual, and Percent Yield

Theoretical yield

• The ___________ amount of product calculated using the balanced equation.

Actual yield

• The ___________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)

To calculate the percent yield, the actual yield and theoretical yield are needed.You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burn and you have to 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

Calculating Percent Yield

Learning CheckWithout 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?

theoretical yield of CO30.0 g O2 x 1 mol O2 x 2 mol CO x 28.01 g CO

32.00 g O2 1 mol O2 1 mol CO= 52.5 g CO (theoretical)

percent yield 40.0 g CO (actual) x 100 = 76.2 % yield

52.5 g CO (theoretical)

Learning CheckWhen 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)

5.00 g H2 x 1 mol H2 x 2 mol NH3 x 17.03 g NH3

2.016 g H2 3 mol H2 1 mol NH3

= 28.2 g NH3 (theoretical)

Percent yield = 16.0 g NH3 x 100 = 56.7 %

28.2 g NH3

Limiting ReactantWhat is the limiting reactant when 2.00g of Na and 2.00g of

Cl2 combine as follows:

2Na + Cl2 2NaCl

Na 22.99 g/mol

Cl 35.45 g/mol

Problem continued….How many grams of the remaining reactant would be left over once the

reaction has run to completion? 2Na + Cl2 2NaCl

Problem continued….If the actual yield of NaCl is 2.29g, what is the percent yield?

Limiting ReactantEthylene burns in air according to the following equation:

C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(l)

a. How many grams of CO2 will be formed when a mixture of 2.93g of C2H4 and 4.29g of O2 combine?

C 12.01 g/molH 1.008 g/molO 16.00 g/mol

Problem continued….b. How many grams of the reactant in excess would be left over once

the reaction has gone to completion?

Problem continued….c. If the percent yield of CO2 is 72.1%, what was the actual yield?

Limiting ReactantSulfur trioxide is prepared from SO2 according to the

following equation: 2SO2 + O2 2SO3

In this reaction, not all SO2 is converted to SO3 even with excess O2. In a given experiment, 21.2g of SO3 is actually produced from 24.0g of SO2. a) What is the theoretical yield of SO3? b) What is the percent yield?

S 32.07 g/mol

O 16.00 g/mol

Chapter 9 Chemical Quantities in Reactions. Mole Relationships in Chemical Equations Copyright ©...

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