chapter 9 chemical quantities. chapter 9 table of contents copyright © cengage learning. all rights...

Chapter 9

Chemical Quantities

Chapter 9

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

9.1 Information Given by Chemical Equations

9.2 Mole–Mole Relationships

9.3 Mass Calculations

9.4 The Concept of Limiting Reactants

9.5 Calculations Involving a Limiting Reactant

9.6Percent Yield

Section 9.1

Information Given by Chemical Equations


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• A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction.

• The coefficients of a balanced equation give the relative numbers of molecules.

Section 9.1



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C2H5OH + 3O2 2CO2 + 3H2O

• The equation is balanced.• All atoms present in the reactants are accounted for in

the products.• 1 molecule of ethanol reacts with 3 molecules of

oxygen to produce 2 molecules of carbon dioxide and 3 molecules of water.

• 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

Section 9.1



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Concept CheckOne of the key steps in producing pure copper from copper ores is the reaction below:

Cu2S(s) + Cu2O(s) → Cu(s) + SO2(g) (unbalanced)

After balancing the reaction, determine how many dozen copper atoms could be produced from a dozen molecules of the sulfide and two dozen of the oxide of copper(I). Also, how many moles of copper could be produced from one mole of the sulfide and two moles of the oxide?

a) 1 dozen and 1 mole

b) 2 dozen and 2 moles

c) 3 dozen and 3 moles

d) 6 dozen and 6 moles

2 6

Section 9.2


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Mole–Mole Relationships

• A balanced equation can predict the moles of product that a given number of moles of reactants will yield.

• 2 mol of H2O yields 2 mol of H2 and 1 mol of O2.

• 4 mol of H2O yields 4 mol of H2 and 2 mol of O2.

2H2O(l) → 2H2(g) + O2(g)

Section 9.2


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• The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation.

Mole Ratio

Section 9.2


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Consider the following balanced equation:

Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)

How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6?

Where are we going?• We want to determine the number of moles of NaF

produced by Na with excess Na2SiF6.

What do we know?• The balanced equation.• We start with 3.50 mol Na.

Example

Section 9.2


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Consider the following balanced equation:

Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)

How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6?

How do we get there?

Example

3.50 mol Na 6 mol NaF

4 mol Na 5.25 mol NaF

Starting

Amount

Mole

Ratio

Section 9.2


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Exercise

Propane, C3H8, is a common fuel used in heating homes in rural areas. Predict how many moles of CO2 are formed when 3.74 moles of propane are burned in excess oxygen according to the equation:

C3H8 + 5O2 → 3CO2 + 4H2O

a) 11.2 moles

b) 7.48 moles

c) 3.74 moles

d) 1.25 moles2

3 8

3 8

3 moles CO3.74 moles C H

1 mole C H

Section 9.3


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Mass Calculations

1. Balance the equation for the reaction.

2. Convert the masses of reactants or products to moles.

3. Use the balanced equation to set up the appropriate mole ratio(s).

4. Use the mole ratio(s) to calculate the number of moles of the desired reactant or product.

5. Convert from moles back to masses.

Steps for Calculating the Masses of Reactants and Products in Chemical Reactions

Section 9.3


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Mass Calculations

• Stoichiometry is the process of using a balanced chemical equation to determine the relative masses of reactants and products involved in a reaction.

Stoichiometry

Section 9.3


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Mass Calculations

For the following unbalanced equation:

Cr(s) + O2(g) → Cr2O3(s)

How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas?

Where are we going?

• We want to determine the mass of Cr2O3 produced by Cr with excess O2.

Example

Section 9.3


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Mass Calculations


Cr(s) + O2(g) → Cr2O3(s)


What do we know?• The unbalanced equation.• We start with 15.0 g Cr.• We know the atomic masses of chromium (52.00

g/mol) and oxygen (16.00 g/mol) from the periodic table.

Example

Section 9.3


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Mass Calculations


Cr(s) + O2(g) → Cr2O3(s)


What do we need to know?• We need to know the balanced equation.

4Cr(s) + 3O2(g) → 2Cr2O3(s)

• We need the molar mass of Cr2O3.

152.00 g/mol

Example

Section 9.3


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Mass Calculations

4Cr(s) + 3O2(g) → 2Cr2O3(s)

• Convert the mass of Cr to moles of Cr.

• Determine the moles of Cr2O3 produced by using the mole ratio from the balanced equation.


1 mol Cr15.0 g Cr = 0.288 mol Cr

52.00 g Cr

2 3

2 3

2 mol Cr O0.288 mol Cr = 0.144 mol Cr O

4 mol Cr

Section 9.3


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Mass Calculations

4Cr(s) + 3O2(g) → 2Cr2O3(s)

• Convert moles of Cr2O3 to grams of Cr2O3.

• Conversion string:


2 3

2 3 2 32 3

152.00 g Cr O0.144 mol Cr O = 21.9 g Cr O

1 mol Cr O

2 3 2 3

2 32 3

2 mol Cr O 152.00 g Cr O1 mol Cr15.0 g Cr = 21.9 g Cr O

52.00 g Cr 4 mol Cr 1 mol Cr O

Section 9.3


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Mass Calculations

C6H6 + H2 C6H12

Reaction Stoichiometry

Balanced?

3

How many grams of cyclohexane C6H12 can be made from 4.5 grams of benzene C6H6?

Use the table method.4.5g

Moles level

Grams level ?

4.5 g C6H6 ---------------- ------------------ ------------------- = 1 mole C6H6 1 mole C6H12 84.174g C6H12

78.1134 g 1 mole C6H6 1 mole C6H12

4.8 g C6H12

How many grams of H2 will be needed?

?

4.5 g C6H6 ---------------- ---------------- ----------------- = 1 mole C6H6 3 moles H2 2.0158g H2

78.1134 g 1 mole C6H6 1 mole H2

.35 g H2

Section 9.3


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Mass CalculationsC6H12O6 + O2 CO2 + H2O6 6 6

How many grams of sugar are needed to make 6.23g water?

6.23 g?

6.23 gw -------------- ------------ ---------------- = 1 molew 1moles 180.1548gs

18.015gw 6 molew 1 moles

10.4 gs

How many grams of carbon dioxide can be produced from 12.84 g of oxygen?

12.84 go -------------- ------------ ------------ = 1 moleo 6 molec 44.009gc

31.998go 6 moleo 1 molec

17.66 gc

12.84 g ?

Section 9.3


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Mass Calculations

Exercise

Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is made according to the following equation:

Sn(s) + 2HF(aq) SnF2(aq) + H2(g)

How many grams of tin(II) fluoride can be made from 55.0 g of hydrogen fluoride if there is plenty of tin available to react?

a) 431 g

b) 215 g

c) 72.6 g

d) 1.37 g

2 2

2

1 mol SnF 156.71 g SnF1 mol HF55.0 g HF

20.008 g HF 2 mol HF 1 mol SnF

Section 9.3


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Mass Calculations

Exercise

Consider the following reaction:

PCl3 + 3H2O H3PO3 + 3HCl

What mass of H2O is needed to completely react with 20.0 g of PCl3?

a) 7.87 g

b) 0.875 g

c) 5.24 g

d) 2.62 g

3 2 2

3

3 3 2

1 mol PCl 3 mol H O 18.016 g H O20.0 g PCl

137.32 g PCl 1 mol PCl 1 mol H O

Section 9.3


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Mass Calculations

Exercise

Consider the following reaction where X represents an unknown element:

6 X(s) + 2 B2O3(s) B4X3(s) + 3 XO2(g)

If 175 g of X reacts with diboron trioxide to produce 2.43 moles of B4X3, what is the identity of X?

a) Ge

b) Mg

c) Si

d) C

4 3

4 3

6 mol X2.43 mol B X 14.6 mol X

1 mol B X

# grams 175 g XMolar Mass = = = 12.0 g/mol

# moles 14.6 mol X

Section 9.4

The Concept of Limiting Reactants


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• Contains the relative amounts of reactants that matches the numbers in the balanced equation.

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

Stoichiometric Mixture

Section 9.4



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

Limiting Reactant Mixture

Section 9.4



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

• Limiting reactant is the reactant that runs out first and thus limits the amounts of product(s) that can form. H2

Limiting Reactant Mixture

Section 9.5

Calculations Involving a Limiting Reactant


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• Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

Section 9.5



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• Methane and water will react to form products according to the equation:

CH4 + H2O 3H2 + CO

Limiting Reactants

Section 9.5



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• H2O molecules are used up first, leaving two CH4 molecules unreacted.

Limiting Reactants

Section 9.5



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• The amount of products that can form is limited by the water.

• Water is the limiting reactant.• Methane is in excess.

Limiting Reactants

Section 9.5



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1. Write and balance the equation for the reaction.

2. Convert known masses of reactants to moles.

3. Using the numbers of moles of reactants and the appropriate mole ratios, determine which reactant is limiting.

4. Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.

5. Convert from moles of product to grams of product, using the molar mass (if this is required by the problem).

Steps for Solving Stoichiometry Problems Involving Limiting Reactants

Section 9.5



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• You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know

in order to determine the mass of product that will be produced?

Example

Section 9.5



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• Where are we going? To determine the mass of product that will be

produced when you react 10.0 g of A with 10.0 g of B.

• What do we need to know? The mole ratio between A, B, and the product

they form. In other words, we need to know the balanced reaction equation.

The molar masses of A, B, and the product they form.

Let’s Think About It

Section 9.5



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• You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:

A + 3B 2C

Example – Continued

Section 9.5



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A + 3B 2C• How do we get there?

Convert known masses of reactants to moles.


1 mol A10.0 g A = 1.00 mol A

10.0 g A

1 mol B10.0 g B = 0.500 mol B

20.0 g B

Section 9.5



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Determine which reactant is limiting.

react with all of the A Only 0.500 mol B is available, so B is the limiting

reactant.


3 mol B1.00 mol A = 3.00 mol B required to

1 mol A

Section 9.5



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Compute the number of moles of C produced.


2 mol C0.500 mol B = 0.333 mol C produced

3 mol B

Section 9.5



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Convert from moles of C to grams of C using the molar mass.


25.0 g C0.333 mol C = 8.33 g C

1 mol C

Section 9.5



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• We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation.

Notice

Section 9.5



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H2 + I2 HI

How many grams of HI can be formed from 2.00 g H2 and 2.00 g of I2?

2

2.00 g2.00 g ?

2.00 g H2 -------------- ------------- ------------------- = 1 mole H2 2 mole HI 127.9124g HI

2.0158 g 1 mole H2 1 mole HI254 g HI

2.00g I2 ------------- --------------- ----------------- = 1 mole I2 2 moles HI 127.9124g HI

253.810 g 1 mole I2 1 mole HI2.02 g HI

Limiting Reactant is always the smallest value!

I2 is the limiting reactant and H2 is in XS.

Section 9.5



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C3H8 + O2 CO2 + H2OHow many grams of CO2 can be formed from 1.44 g C3H8

and 2.65 g of O2?

3

2.65g1.44 g ?

1.44 g C3H8 ---------------- ---------------- ------------------- = 1 mole C3H8 3 mole CO2 44.009 g CO2

44.0962 g 1 mole C3H8 1 mole CO2

4.31 g

CO2

2.65g O2 ------------- ----------------- ------------------- = 1 mole O2 3 moles CO2 44.009 g CO2

31.998 g 5 mole O2 1 mole CO2

2.19 g CO2

45

O2 is the limiting reactant and C3H8 is in XS.

Section 9.5



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Concept Check

Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation:

2H2 + O2 2H2O

a) 2 moles of H2 and 2 moles of O2

b) 2 moles of H2 and 3 moles of O2

c) 2 moles of H2 and 1 mole of O2

d) 3 moles of H2 and 1 mole of O2

e) Each produce the same amount of product.

Section 9.5



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Concept Check

Consider the equation: A + 3B 4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete?

a) A is the leftover reactant because you only need 2 moles of A and have 3.

b) A is the leftover reactant because for every 1 mole of A, 4 moles of C are produced.

c) B is the leftover reactant because you have more moles of B than A.

d) B is the leftover reactant because 3 moles of B react with every 1 mole of A.

Section 9.5



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Exercise

How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process?

N2 + 3H2 → 2NH3

a) 2.00 g

b) 3.00 g

c) 3.64 g

d) 1.82 g

3.00 g N2(1mole/28g N2)(2 moles NH3/1 moles N2)

(17 g NH3/1 mole NH3) = 3.64 g NH3

3.00 g H2(1mole/2g H2)(2 moles NH3/3 moles H2)

(17 g NH3/1 mole NH3) = 17.0 g NH3

Section 9.6

Percent Yield


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• An important indicator of the efficiency of a particular laboratory or industrial reaction.

Percent Yield

Section 9.6

Percent Yield


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• Theoretical Yield The maximum amount of a given

product that can be formed when the limiting reactant is completely consumed.

• The actual yield (amount actually produced) of a reaction is usually less than the maximum expected (theoretical yield).

Section 9.6

Percent Yield


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• The actual amount of a given product as the percentage of the theoretical yield.

Percent Yield

yieldpercent%100 yieldlTheoretica

yieldActual

Section 9.6

Percent Yield


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A + 3B 2C

We determined that 8.33 g C should be produced.

7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?

• Where are we going? We want to determine the percent yield of the

reaction.

Example – Recall

Section 9.6

Percent Yield


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A + 3B 2C

We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?

• What do we know? We know the actual and theoretical yields.

Example – Recall

Section 9.6

Percent Yield


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A + 3B 2C

We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?

• How do we get there?

Example – Recall

7.23 g C 100% = 86.8%

8.33 g C

Section 9.6

Percent Yield


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Exercise

Find the percent yield of product if 1.50 g of SO3 is produced from 1.00 g of O2 and excess sulfur via the reaction:

2S + 3O2 → 2SO3

a) 40.0%

b) 80.0%

c) 89.8%

d) 92.4%

Section 9.6

Percent Yield


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Exercise

Consider the following reaction:

P4(s) + 6F2(g) 4PF3(g)

What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield?

a) 29.9 g

b) 46.1 g

c) 184 g

d) 738 g

Section 9.6

Percent Yield


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Section 9.6

Percent Yield


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Assignment: Read 5.6 up to sample (238-9)1. Define the following terms: yield, theoretical yield, actual yield,

percentage yield.2. Based on your reading, give 4 reasons why the actual yield in a

chemical reaction often falls short of the theoretical yield.3. Read the sample problem on the next slide and try the practice

problem on slide number 54. When 5.00 g of KClO3 is heated it decomposes according to the

equation: 2KClO3 2KCl + 3O2

a) Calculate the theoretical yield of oxygen.b) Give the % yield if 1.78 g of O2 is produced.

c) How much O2 would be produced if the percentage yield was 78.5%?

Section 9.6

Percent Yield


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Answers

1) Yield: the amount of productTheoretical yield: the amount of product we expect, based on

stoichiometric calculationsActual yield: amount of product from a procedure or experiment

(this is given in the question)

Percent yield: = x 100%actual yield

theoretical yield

2)• Not all product is recovered (e.g. spattering)• Reactant impurities (e.g. weigh out 100 g of chemical which

has 20 g of junk)• A side reaction occurs (e.g. MgO vs. Mg3N2)• The reaction does not go to completion

Section 9.6

Percent Yield


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Sample problem

Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?

Step 1: write the balanced chemical equation2H2 + O2 2H2O

Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:

2 mol H2O

2 mol H2

x # g H2O= 16 g H2 143 g= 18.02 g H2O

1 mol H2Ox

1 mol H2

2.02 g H2

x

Step 3: Calculate % yield

138 g H2O

143 g H2O= % yield = x 100% 96.7%=

actual

theoretical x 100%

Section 9.6

Percent Yield


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Practice problem

Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?

Step 1: write the balanced chemical equationN2 + 3H2 2NH3

Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:

2 mol NH3

3 mol H2

x # g NH3= 20.0 mol H2 227 g= 17.04 g NH3

1 mol NH3

x

Step 3: Calculate % yield

40.5 g NH3

227 g NH3

= % yield = x 100% 17.8%= actual

theoretical x 100%

Section 9.6

Percent Yield


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4) 2KClO3 2KCl + 3O2

a)

b)

c)

Answers

3 mol O2

2 mol KClO3

x

# g O2= (also works if you use mol O2)

5.00 g KClO3

1.958 g=

32 g O2

1 mol O2

x 1 mol KClO3

122.55 g KClO3

x

1.78 g O2

1.958 g O2

= % yield = x 100% 90.9%= actual

theoretical x 100%

x g O2

1.958 g O2

= % yield = x 100% 78.5%= actual

theoretical x 100%

x g O278.5% x 1.958 g O2

100%= 1.537 g O2=

Section 9.6

Percent Yield


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Challenging question

2H2 + O2 2H2O

What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?

Hint: determine limiting reagent first

2 mol H2O

2 mol H2

x # g H2O= 7.0 g H2 62.4 g= 18.02 g H2O

1 mol H2Ox

1 mol H2

2.02 g H2

x

58 g H2O

62.4 g H2O= % yield = x 100% 92.9%=

actual

theoretical x 100%

2 mol H2O

1 mol O2

x # g H2O= 60 g O2 68 g= 18.02 g H2O

1 mol H2Ox

1 mol O2

32 g O2

x

Section 9.6

Percent Yield


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More Percent Yield QuestionsNote: try “shortcut” for limiting reagent problems

1. The electrolysis of water forms H2 and O2.

2H2O 2H2 + O2 What is the % yield of O2 if 12.3 g

of O2 is produced from the decomposition of 14.0 g H2O?

2. 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3 2KCI + 3O2

3. What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S FeS

Section 9.6

Percent Yield


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More Percent Yield Questions4. Iron pyrites (FeS2) reacts with oxygen according to the following

equation:

4FeS2 + 11O2 2Fe2O3 + 8SO2

If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric

oxide is produced. What is the percent yield of ferric oxide?

5. 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this

reaction if the % yield for the process is 42%?MnO2 + 4HCI MnCl2 + 2H2O + Cl2

Section 9.6

Percent Yield


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Q1

1. The electrolysis of water forms H2 & O2. 2H2O 2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O?

• Actual yield is given: 12.3 g O2

• Next, calculate theoretical yield

1 mol O2

2 mol H2Ox

# g O2=

14.0 g H2O 12.43 g= 32 g O2

1 mol O2

x 1 mol H2O

18.02 g H2Ox

Finally, calculate % yield

12.3 g O2

12.43 g O2

= % yield = x 100% 98.9%= actual

theoretical x 100%

Section 9.6

Percent Yield


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Q2

2. 107 g of oxygen is produced by heating 300 grams of potassium chlorate.

2KClO3 2KCI + 3O2 • Actual yield is given: 107 g O2

• Next, calculate theoretical yield

3 mol O2

2 mol KClO3

x

# g O2=

300 g KClO3

117.5 g=

32 g O2

1 mol O2

x 1 mol KClO3

122.55 g KClO3

x


107 g O2

117.5 g O2

= % yield = x 100% 91.1%= actual

theoretical x 100%

Section 9.6

Percent Yield


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Q3

3. What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide?

Fe + S FeS• Actual yield is given: 220 g FeS• Next, calculate theoretical yield

1 mol FeS

1 mol Fex # g FeS= 3.00 mol Fe 263.7 g=

87.91 g FeS

1 mol FeSx


220 g O2

263.7 g O2

= % yield = x 100% 83.4%= actual

theoretical x 100%

Section 9.6

Percent Yield


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4. 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3?

First, determine limiting reagent

2 mol Fe2O3

11 mol O2 x 200 g O2

181.48 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol O2

32 g O2 x

2 mol Fe2O3

4 mol FeS2 x

# g Fe2O3=

300 g FeS2

199.7 g Fe2O3=

159.7 g Fe2O3

1 mol Fe2O3

x 1 mol FeS2

119.97 g FeS2 x

143 g Fe2O3

181.48 g Fe2O3

= % yield = x 100% 78.8%= actual

theoretical x 100%

Section 9.6

Percent Yield


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5. 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI MnCl2 + 2H2O + Cl2

1 mol Cl2

4 mol HClx 3.5 mol HCl 62.13 g Cl2=

71 g Cl2

1 mol Cl2

x

1 mol Cl2

1 mol MnO2 x

# g Cl2=

70 g MnO2

57.08 g Cl2=

70.9 g Cl2

1 mol Cl2

x 1 mol MnO2

86.94 g MnO2 x

x g Cl2

57.08 g Cl2

= % yield = x 100% 42%= actual

theoretical x 100%

x g Cl242% x 57.08 g Cl2

100%= 24 g Cl2=

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