CHEM 100 Fall 2013. chapter 4 -1. Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30

CHEM 100 Fall 2013 . chapter 4 -1 .

Instructor: Dr. Upali Siriwardanee-mail: [email protected]: CTH 311 Phone 257-4941Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m Tu,Th,F 8:00 - 10:00 a.m. Or by appointmentTest Dates:

Chemistry 100(02) Fall 2013

September 30, 2013 (Test 1): Chapter 1 & 2 October 21, 2013 (Test 2): Chapter 3 & 4 November 13, 2013 (Test 3) Chapter 5 & 6November 14, 2013 (Make-up test) comprehensive: Chapters 1-6 9:30-10:45:15 AM, CTH 328


REQUIRED :Textbook: Principles of Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also purchase the Mastering Chemistry Group Homework, Slides and Exam review guides and sample exam questions are available online: http://moodle.latech.edu/ and follow the course information links.OPTIONAL : Study Guide: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd EditionStudent Solutions Manual: Chemistry: A Molecular Approach, 2nd Edition-Nivaldo J. Tro 2nd

Text Book & Resources

http://moodle.latech.edu/


4.1 Global Warming and the Combustion of Fossil Fuels…………………. 1274.2 Reaction Stoichiometry: How Much Carbon Dioxide?......................... 1284.3 Limiting Reactant, Theoretical Yield, and Percent Yield………………. 1334.4 Solution Concentration and Solution Stoichiometry………………….. 1404.5 Types of Aqueous Solutions and Solubility…………………………….. 1464.6 Precipitation Reactions………………………………………………….. 1504.7 Representing Aqueous Reactions: Molecular, Ionic, and Complete Ionic Equations…………………………………………………………………........ 1534.8 Acid–Base and Gas-Evolution Reactions……………………………..... 1554.9 Oxidation–Reduction Reactions…………………………………………. 162

Chapter 4. Chemical Quantities and Aqueous Reactions


Chapter 4. Chemical Quantities and Aqueous Reactions

•Global Warming •Stoichiometry •Reactions in One reactant in Limited Supply•Limiting reactant•Evaluating Success of Synthesis •Theoretical Yield•Actual Yield•Percent Yield•Solution Concentration•Molecular, Ionic, and Complete Ionic Equations

•Precipitation Reactions•Acid–Base and Gas-Evolution Reactions•Oxidation–Reduction Reactions•Stoichiometry in Solution Reactions


What is Global Warming?


Reaction Stoichiometry

• According to the Law of Conversion of Matter:– Matter is neither created nor destroyed in a chemical reaction.– A balanced chemical equation illustrates the law of

conversation of matter.• In a balanced reaction:

– Total mass of reactants = Total mass of products– A balanced reaction has the same type and quantity of

atoms on both sides of the reaction.

• Stoichiometry is based on the law of conversion of matter.– Stoichiometry studies the quantitative aspects of chemical

reactions.


Reaction Stoichiometry: What it means

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

This equation means:

4 atoms Fe + 3 molecules O2 2 molecules

Fe2O3

4 moles Fe + 3 moles O2 2 moles

Fe2O3

4 moles Fe atoms = Fe atoms and 6 moles O atoms and 6 mole O atoms

23.4 g Fe + 96.0 g O2 = 415.4 g Fe2O3

415.4 g = 415.4 g of reactants of products


1) Given the balanced reactions:

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

Write the mole conversion factors:


1. Need a balanced reaction to determine the stoichiometric relationship between:

a. Reactants and reactants orb. Reactants and products orc. Products and products

2. Go to the mole:d. If mass is given, then divide by molecular mass:

mass (g) /mol. mass (g/mole) = mole

b. If volume and molarity (M) are given, then:Ma x Va = moles A

3. Use the stoichiometric factor to convert from mole A to mole B to solve problem.

Strategy Behind Solving STOICHIOMETRY Problems


10

Limiting Reactant

• For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others.

• When this reactant is used up, the reaction stops and no more product is made.

• The reactant that limits the amount of product is called the limiting reactant (limiting reagent).– The limiting reactant is completely consumed.

• Reactants not completely consumed are called excess reactants.

• The amount of product that can be made from the limiting reactant is called the theoretical yield.


Limiting Reactant


Analogy in Recipe : Making Cheese Sandwiches

You were given 20 slices bread, 5 slices of cheese, 4 slices of ham If you want to make sandwiches containing two slices bread and one slice of cheese and one slice of ham

How many sandwiches you could make? What is the limiting ingredient?


• Theoretical Yield and Limiting Reagent (Reactants)• Percent Yield

Using Stoichiometry to Predict


Problem:

The following unbalanced equation is the chemical reaction

associated with photosynthesis.

CO2(g) + H2O(l) C6H6O6(s) + O2(g)

With adequate water, suppose a plant consumes 37.8 grams of CO2

during the week.

Determine how many grams of glucose (C6H6O6) would be produced

by the plant during this one-week period.


Problem Strategy:

Determine how many grams of glucose (C6H6O6) would be produced

by the plant if 37.8 grams of CO2 is consumed.

1. Need a balanced equation:

6 CO2(g) + 6 H2O(l) C6H6O6(s) + 6 O2(g)

2. Determine moles of CO2 consumed.

37.8 g CO2 x (1 mol CO2/44.0 g) = 0.859 mol CO2

3. Determine how many moles of C6H6O6 would be produced. Use the stoichiometric

relationship between CO2 and C6H6O6. In this reaction the relationship is 6 mole CO2 to 1 mole C6H6O6.

0.859 mol CO2 x (1 mol C6H6O6/6 mol CO2) = 0.143 mol C6H6O6

4. Determine the grams of C6H6O6 produced.

0.143 mol C6H6O6 x (1 mol / 180.0 g/1 mol C6H6O6) = 25.8 g C6H6O6


2) My recipe ratio for a bacon double cheeseburger is:1 bun + 2 hamburger patties + 1 slices of cheese + 2 strips of bacon

If you start with: 2 bun, 6 patties, 4 slices of cheese, 6 strips of bacon a) How many bacon double cheeseburgers can you make?b) Which ingredient is limiting?c) What ingredients would be left over or in excess?

hamburger bun hamburger patties slices of cheese strips of bacon

1 2 1 2

2 6 4 6


3) Determine if the reaction is stoichiometric, or one reagent is in excess/one reagent is limiting for the reaction:

3H2(g) + 1N2(g) = 2 NH3(g)

a) If 3 mole H2 and 1.00 mole of N2 are reacted according to the equation:

Stoichiometric ratio (mole H2/ mole of N2)=

Actual mole ratio (mole H2/ mole of N2) =

b) Is stoichiometric ratio equal, grate or less than actual mole ratio?

c) Is the reaction stoichiometric, H2 limiting or N2 limiting?


4) Determine if the reaction is stoichiometric, or one reagent is in excess/one reagent is limiting for the reaction: reaction:

2H2(g) + 1O2(g) = 2 H2O (l)

a) If 1 mole H2 and 2.00 mole of O2 are reacted according to the equation:

Stoichiometric ratio (mole H2/ mole of O 2)=

Actual mole ratio (mole H2/ mole of O 2) =

b) Is stoichiometric equal, grate or less than actual mole ratio?

c) Is the reaction stochiometric, H2 limiting or O 2

limiting?


COMPARE THEORETICAL YIELDS TO DETERMINE THE LIMITING REACTANT:

• If all 8.10 g O2 were used, then 17.2 g of Al2O3 would be produced.

• If all 5.40 g Al were used, then 10.2 g of Al2O3 would be produced.

10.2 g < 17.2 g

• The limiting reactant is Al.

• Theoretical yield is 10.2 g Al2O3.

To determine the percent yield of the reaction:

(4.50 g/10.2) x 100 = 44.1%44.1% is the percent yield for this reaction.


Which reactant will remain when the reaction is complete?

• Al was the limiting reactant.

– Therefore, O2 was in excess. But by how much?

– First find how much oxygen gas was required.

– Then find how much oxygen gas is in excess.


4 Al + 3 O2 products

0.200 mol = LR0.200 mol = LR 0.253 mol0.253 mol

O2 available – O2 required = excess O2

0.253 mol O2 - 0.150 mol O2 = 0.103 mol O2 left over

= 0.103 mol O2 in excess, or 3.30 grams O2

0.200 mol Al x (3 mol O2/4 mol Al) = 0.15 mol O2

0.150 mol of O2 is required to react with all 0.200 mol of Al.

How to Determine the Amount of Excess Reagent Left Over


5) 2 Al (s) + Fe2O3 (s) = 2 Fe(l) + Al2O3

takes place in the thermite mixture when it is ignited by a magnesium ribbon. A thermite mixture contains a mass ratio of 1 to 2 for Al and Fe2O3. Which one is the limiting reagent?


6) Consider the reaction: 2H2(g) + O2(g) = 2 H2O(l)

Equal weights of H2 and O2 are placed in a balloon and then ignited. Assume reaction goes to completion, which gas is the excess reagent?a) How many moles of H2O will be produced by 0.80 mole of O2 and excess H2, according to the equation?

b) How many moles of H2O will be produced by 25.6g of O2 and excess H2, according to the equation?

c) How many grams moles of H2O will be produced by 25.6g of O2 and excess H2, according to the equation?


7) Two moles of Mg and five moles of O2 are placed in a reaction vessel, and then the Mg is ignited to produce MgO(F.W.=40.31 g/mole)

a) The balanced chemical reaction:

b) The limiting reactant?

c) Actual mole ratio:

d) Calculated mole ratio:

e) How many moles of MgO are formed?

f) What is the weight of MgO formed?


9) Write the balanced chemical equation for the reaction where zinc is producing silver in a single displacement reaction with silver chloride.


Theoretical and Percent Yield

• Theoretical Yield– Predicting what could/should/would be produced

from a given amount of reactant(s)• Limiting reagent (reactant)

– The determining reactant

– Using stoichiometric relationship(s) from balanced reaction

• Percent Yield– Actual amount produced– % yield = (actual yield/theoretical yield) x 100


Strategy:

1. Check to see if reaction given is balanced.

2. Determine limiting reactant and theoretical yield.

Solution:

1. Balance reaction: 2 Mg(s) + O2(g) 2 MgO(s)

2. Determine limiting reactant * theoretical yield.


Predicting Theoretical Yield

Problem: Determine the theoretical yield for the following reaction:

Mg(s) + O2(g) MgO(s)

when 42.5 g Mg(s) and 33.8 g O2(g) are reacted.


10) If 15.00 g of Zn (A.M.= 65.39 g/mole) reacts with silver with 25.00 g AgCl (F.M.= 143.35 g/mole)?

a) Actual moles of Zn= 0.229

b) Actual moles of AgCl= 0.174

c) Stoichiometric ratio (mole Zn/ mole of AgCl)= ½ = 0.5

d) Actual mole ratio (mole Zn/ mole of AgCl)= 1.32

e) What is limiting reactant?

b) Mole of Ag produced?

c) Grams of Ag produced (theoretical yield)?

d) If the actual yield of Ag was found to be 16.1 g Ag,

the percent yield =


8) When 10.0 g pure solid calcium carbonate, CaCO3 is heated and converted to solid calcium oxide CaO and CO2 gas:

a) The balanced chemical reaction:

b) How much calcium oxide should be theoretically obtained?

c) If 4.20 g of CaO is actually produced what is the percent yield of CaO?


Problem: Ammonium nitrate (NH4NO3) decomposes

to N2O(g) and H2O(g).

1. Predict how many grams of H2O will be produced from the decomposition of 0.454 kg of NH4NO3.

2. If 131 grams of water is produced, what is the % yield for this reaction?

Strategy: 1. Write a balanced reaction.

2. Determine the theoretical yield.

3. Calculate the percent (%) yield.


Problem Solution: Ammonium nitrate (NH4NO3) decomposes


Balance equation: NH4NO3(s) N2O(g) + 2 H2O(g)

1. Predict how many grams of H2O will be produced from the decomposition of 0.454 kg of NH4NO3.

0.454 kg NH4NO3 x (1000 g/1 kg) x (1 mol NH4NO3/80.04 g)

= 5.68 mol NH4NO3

5.68 mol NH4NO3 x (2 mol H2O/1mol NH4NO3 ) = 11.4 mol H2O

11.4 mol H2O x (18.0 g/1 mol H2O) = 204 grams H2O

204 grams of H2O is the predicted or theoretical yield.


Problem: Ammonium nitrate (NH4NO3) decomposes


Balance equation: NH4NO3(s) N2O(g) + 2 H2O(g)

If 131 grams of water is produced, what is the % yield for this reaction?

% yield = (131 g/204 g) x 100 = 64.2%

62.4% is the percent yield for the reaction.


Problem:

When 28.6 kg of carbon reacted with 88.2 kg of TiO2, 42.8 kg of Ti was obtained.

Reaction: TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

Strategy: 1. Determine limiting reactant.

2. Calculate theoretical yield.

3. Calculate percent yield.


Problem Solution: When 28.6 kg of carbon reacted with 88.2 kg of TiO2, 42.8 kg of Ti was obtained.

Reaction: TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) It is balanced!

1. Determine limiting reactant.

28.6 kg carbon: 28.6 kg C x (1000 g/1 kg) = 2.86 x 104 g C

88.2 kg TiO2 88.2 kg TiO2 x (1000 g/1 kg) = 8.82 x 104 g TiO2

2.86 x 104 g C x (1 mol C/12.0 g) x (1 mol Ti/2 mol C)

= 1.19 x 103 mol Ti THIS IS THE MAXIMUM MOLE OF TiTHAT CAN BE PRODUCED IF ALL CARBON IS USED.

8.82 x 104 g TiO2 x (1 mol TiO2/79.9 g) x (1 mol Ti/1 mol TiO2)

= 1.10 x 103 mol Ti THIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL TiO2 IS USED.




2. Determining the theoretical yield from limiting reactant.

= 1.19 x 103 mol Ti THIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL CARBON IS USED.

= 1.10 x 103 mol Ti THIS IS THE MAXIMUM MOLE OF Ti THAT CAN BE PRODUCED IF ALL TiO2 IS USED.

1.10 x 103 < 1.19 x 103; THEREFORE, the limiting reagent is TIO2 and the theoretical yield in moles of Ti is 1.10 x 103.




3. Determining percent yield:

The theoretical yield in moles of Ti is 1.10 x 103.

1.10 x 103 mol Ti x (47.9 g/1 mol Ti) = 5.29 x 104 g Ti

42.8 kg is actual yield of Ti.

42.8 kg x (1000 g/1 kg) = 4.28 x 104 g Ti was produced.

(4.28 x 104 g Ti/5.29 x 104 g Ti) x 100 = 80.9%

80.9% is the percent yield for this reaction.


PROBLEM: Given the following chemical reaction:Al(s) + O2(g) Al2O3(s)

1. Determine how many grams of Al2O3 can form when

5.40 grams of Al and 8.10 grams of O2 are reacted.

2. If 4.50 grams of Al2O3 was produced, what is the

percent yield for this reaction?

3. Which reactant was in excess and how much (grams)

of this reactant remained after the reaction came to

completion?


Step 1: Balance the reaction and calculate theoretical yield for each reactant.

Balanced reaction: 4 Al(s) + 3 O2(g) 2 Al2O3(s)

5.40 g Al x (1 mol/27.0 g Al) = 0.200 mol Al

0.200 mol Al x 2 mol Al2O3 = 0.100 mol Al2O3

4 mol Al

stoichiometric factor

0.100 mol Al2O3 x 101.96 g = 10.2 g Al2O3

1 mol Al2O3

This means that if all 5.40 g Al were consumed, then only 10.2 grams of Al2O3 COULD be produced.

NOW DETERMINE THE THEORITICAL YIELD IF ALL 8.10 grams of O2 WERE USED.


Step 1: Balance the reaction and calculate theoretical yield for each reactant.

Balanced reaction: 4 Al(s) + 3 O2(g) 2 Al2O3(s)

8.10 g O2 x ( 1 mol/32.00 g) = 0.253 mol O2

0.253 mol O2 x 2 mol Al2O3 = 0.169 mol Al2O3

3 mol O2

stoichiometric factor

0.169 mol Al2O3 x 101.96 g = 17.2 g Al2O3

1 mol Al2O3

This means that if all 8.10 grams of O2 were consumed, then 17.2 grams is the MOST that COULD be produced.


PROBLEM: Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn metal, what volume of 2.50 M HCl is needed to covert the Zn metal completely to Zn2+ ions?

Step 1: Write the balanced equation.

Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g)

Step 2: Calculate amount of Zn.

10.0 g Zn x (1 mol Zn/65.39 g) = 0.153 mol Zn

Step 3: Use the stoichiometric factor.

0.153 mol Zn x (2 mol HCl/1 mol Zn) = 0.306 mol HCl

Step 4: Calculate volume of HCl required.

0.306 mol HCl x (1.00 L/2.50 mol) = 0.122 L HCl


Problem: 76.80 grams of apple juice (malic acid) requires 34.56 mL

of 0.664 M NaOH to reach the endpoint in a titration.

What is the (w/w)% of malic acid in this

sample?

Reaction: C4H6O5(aq) + 2 NaOH(aq) --> Na2C4H4O5(aq) + 2 H2O(l)

(malic acid)

Step 1: Calculate amount of NaOH used.

M x V = (0.664 M)(0.03456 L) = 0.0229 mol NaOH

Step 2: Calculate amount of malic acid titrated.

0.0229 mol NaOH x (1 mol malic acid/2 mol NaOH)

= 0.0115 mol malic acid

Step 3: Calculate the grams of malic acid in 0.0115 moles.

0.0115 mol malic acid x (134 g/1 mol malic acid) = 1.54 g

Step 4: Calculate the weight (w/w)% malic acid in apple sample.

(1.54 g/76.80) x 100 = 2.01 (w/w)%


Chemical Analysis Problem:

An impure sample of the mineral contains Na2SO4.

The mass of mineral sample = 0.123 g

All of the Na2SO4 in the sample is converted to insoluble BaSO4.

The mass of BaSO4 is 0.177 g.

Determine the mass percent of Na2SO4 in the mineral.


Answer:

Reaction equations:

Na2SO4(aq) + BaCl2(aq) 2 NaCl(aq) + BaSO4(s)

0.177 g BaSO4 x (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4

7.58 x 10-4 mol BaSO4 x (1 mol Na2SO4/1 mol BaSO4) = 7.58 x 10-4 mol Na2SO4

7.58 x 10-4 mol Na2SO4 x (142.0 g/1 mol) = 0.108 g Na2SO4

(0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4

Documents

CHEM 100 Fall 2013. chapter 4 -1. Instructor: Dr. Upali Siriwardane e-mail: [email protected] Office: CTH 311 Phone 257-4941 Office Hours: M,W, 8:00-9:30