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Chemistry II Stoichiometry Chapter 3
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Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Dec 29, 2015

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Page 1: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Chemistry II

Stoichiometry

Chapter 3

Page 2: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Stoichiometery: Calculations with

Chemical Formulas and Equations

Page 3: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

1. ) What are the advantages of writing chemical formulas over writing chemical names?

•abbreviation•important quantitative information•Structural information

Page 4: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

2.)What two things is stoichiometry dependent on?

1.)Our understanding of atomic masses and 2.the Law of Conservation of mass

Page 5: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

3.) How are chemical reactions represented?

By chemical equations

Page 6: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

4.) In a chemical reaction the starting substances are called reactants and the substances produced are called products.

Page 7: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

5.) The number written before the formula is called the coefficient and the number after the symbol used to indicate the number of atoms of that element in the formula is called the subscript.

Page 8: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

6.) When balancing equations which number can we change?

We change the coefficient but not the subscript. Changing the coefficient changes the number of particles. Changing the subscripts changes the identity of the substance.

Page 9: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

7.) How do we indicate physical state of reactants and products?

By using the following symbols:

Gas (g) solid (s)

Liquid (l) aqueous (aq)

Page 10: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

8.) How do we indicate that the addition of heat is necessary?

By placing the delta symbol, Δ, above the arrow.

Page 11: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Balance this equation:Na(s) + H2O(l) → NaOH(aq) + H2(g)

Page 12: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Balance this equation:2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

Page 13: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Balance the following:

Fe + O2 → Fe2O3

C2H4 + O2 → CO2 + H2O

Page 14: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Balance the following:

4Fe + 3O2 → 2Fe2O3

C2H4 + 3O2 → 2CO2 + 2H2O

Page 15: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 2: Some Simple Patterns

In this section we will review the five basic classifications of chemical reactions. Not all chemical reactions will fit neatly into one of the five classes. But, by recognizing typical patterns you will be more likely to predict products.

Page 16: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

9.) What are the five common types of reactions?

Combination (synthesis)DecompositionSingle replacementDouble replacementcombustion

Page 17: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

10.) What is a combination reaction?

In a combination reaction, two or more substances react to form a single product.

A + B → AB2Mg + O2→ 2 MgO

Page 18: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

11.) What is produced by a direct combination reaction between a metal and a nonmetal?

An ionic solid

2 Li + Cl2 → 2LiCl

Page 19: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

12.)What is a decomposition reaction?

A reaction in which one substance produces two or more other substances.

CaCO3 → CaO + CO2

2HgO → 2 Hg + O2

Page 20: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

13.)What is a single replacement reaction?

An uncombined element displaces an element that is part of a compound. The reactants in a single replacement reaction are always one compound and one element.

Page 21: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

13.)What is a single replacement reaction? contd

A + BX → AX + BMg + 2HCl → MgCl2 + H2

Page 22: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

In a single replacement reaction, metals replace metals and nonmetals replace nonmetals.

The activity series can be used to determine if a single replacement reaction will happen or not.

Page 23: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

14.)Describe a double replacement reaction.

In a double replacement reactions, atoms or ions from two different compounds replace each other.

Page 24: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

14.Describe a double replacement reaction. contd

A double replacement reaction will have two compounds as reactants and two compounds as products.

AX + BY → AY + BXCaCO3 + 2HCl → CaCl2 + H2CO3

Page 25: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Although most chemical reactions will fit into one of three categories, there are many that do not. Combustion reactions often do not fit into one of these categories.

Page 26: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

15.) What are combustion reactions?

Rapid reactions that produce flames. Combustion reactions include O2 as a reactant.

When hydrocarbons and oxygen containing derivatives combust, they form CO2 and H2O as products.

Page 27: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The combination reaction that occurs when lithium metal and fluorine gas react.

Page 28: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The combination reaction that occurs when lithium metal and fluorine gas react.

2Li + F2 → 2LiF

Page 29: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The decomposition reaction that occurs when solid barium carbonate is heated. (Two products form: a solid and a gas.)

Page 30: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The decomposition reaction that occurs when solid barium carbonate is heated. (Two products form: a solid and a gas.)

BaCO3 → BaO + CO2

Page 31: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

Solid mercury(II) sulfide decomposes into its component elements when heated.

Page 32: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

Solid mercury(II) sulfide decomposes into its component elements when heated.

HgS → Hg + S

Page 33: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The surface of aluminum metal undergoes a combination reaction with oxygen in the air.

Page 34: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

The surface of aluminum metal undergoes a combination reaction with oxygen in the air.

4Al + 3O2 → 2Al2O3

Page 35: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

methanol, CH3OH(l) , is burned in air.

Page 36: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Write balanced equations for the following reactions:

methanol, CH3OH(l) , is burned in air.

2CH3OH + 3O2 → 2CO2 + 4H2O

Page 37: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

16.)What is formula weight?

The sum of the atomic weights of each atom in its chemical formula.

Formula weights are given in a.m.u.’s

Page 38: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the formula weight of sucrose C12H22O11

Page 39: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the formula weight of sucrose C12H22O11

342.0 amu

Page 40: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the formula weight of Ca(NO3)2

Page 41: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the formula weight of Ca(NO3)2

164.1 amu

Page 42: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

17.)What is molecular weight?

If the chemical formula is a molecule (covalently bonded) we call the formula weight “molecular weight.”

Page 43: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

18.) What are formula units?

Formula units represent the chemical formula of ionic compounds.

We speak of molecules for molecular compounds and formula units for ionic compounds.

Page 44: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

19.) What is percent composition?

The percent by mass contributed by each element in a substance.

Page 45: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

20.) How do you calculate the % composition ?

% element= (# atoms)( atomic weight) x 100

formula weight of compound

Page 46: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the % composition of C12H22O11

%C = (12)(12 amu) X 100 = 42.1%

342.0 amu

% H = (22)(1 amu) x 100 = 6.4%

342.0 amu

%O = (11)(16.0 amu) x 100 = 51.5%

342.0 amu

Page 47: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the percentage of nitrogen in Ca(NO3)2

Page 48: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the percentage of nitrogen in Ca(NO3)2

17.1 %

Page 49: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 4The mole

# atomsmass mass Ratio of

of H of O masses

1 1 161:16

Page 50: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 4The mole

# atomsmass mass Ratio of

of H of O masses

1 1 16 1:16

2 2 32 1:16

Page 51: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 4The mole

# atoms mass mass Ratio of of H of O masses

1 1 16 1:162 2 32 1:16

10 10 160 1:16

Page 52: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

As long as the # of atoms are the same then the ratio of masses is the same as the reported atomic masses.

Page 53: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

21.) Who was Amedeo Avogadro?

An Italian Chemist who determined that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Page 54: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

22.)What is the mole?

A mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.0 grams of carbon-12.

Page 55: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

23.)What is Avogadro’s number?

6.02 x 1023

1 mole of carbon 12 has 6.02 x 1023 carbon 12 atoms.

Page 56: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is Avogadro’s number?

6.02 x 1023

1 mole of H2O has 6.02 x 1023 water molecules.

Page 57: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is Avogadro’s number?

6.02 x 1023

1 mole of NaCl has 6.02 x 1023 formula units

Page 58: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Without using a calculator, arrange the following samples in order of increasing numbers of carbon atoms:

12 g 12C, 1 mol C2H2, 9x1023molecules of CO2

Page 59: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

12 g 12C < 9x1023 CO2< 1 mole C2H2

Page 60: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the number of H atoms in 0.350 mol of C6H12O6

Page 61: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the number of H atoms in 0.350 mol of C6H12O6

(0.350 mole C6H12O6 )(12 mole H/ 1 C6H12O6 )

(6.02 x 1023 H atoms/ 1 mole H atoms)= 2.53 x 1024 mole H

Page 62: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atoms are in 0.25 mole Ca(NO3)2?

Page 63: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atoms are in 0.25 mole Ca(NO3)2?

9.0 x 10 23

Page 64: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atom are in 1.50 mole of sodium carbonate?

Page 65: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atom are in 1.50 mole of sodium carbonate?

2.71 x 1024

Page 66: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Molar Mass A mole of any two objects will be the same

number, but will have different masses. When we are dealing with a particular isotope

of an element we use the mass of the isotope for molar mass. ( 1 mole of Carbon-12 is 12g/mol.) Otherwise, we use the atomic weight listed on the periodic table for carbon 12.0107 g/mole.

Page 67: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

24What is molar mass?

The mass in grams of 1 mole of a substance. The molar mass in g/mole is numerically equal to its formula weight.

Page 68: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

A single formula unit of NaCl would be 58 amu( formula mass.)

A mole of NaCl formula units would be 58 gram ( molar mass)

Page 69: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 1.000 moles of glucose C6H12O6?

Page 70: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 1.000 moles of glucose C6H12O6?

(6) (12) + (12) (1) + (6)(16) = 180 g/mole

Page 71: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the number of moles of glucose (C6H12O6) in 5.380 g C6H12O6

Page 72: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the number of moles of glucose (C6H12O6) in 5.380 g C6H12O6

5.380 g (1 mole C6H12O6 / 180 g) = 0.02989 mole

Page 73: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

The mole provides the bridge between mass, number and volume.

diagram

Page 74: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

25.) What is the volume of any gas at STP?

1 mole of any gas will occupy a volume of 22.4L

Page 75: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the mass in grams of 0.433 mole of calcium nitrate.

Page 76: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Calculate the mass in grams of 0.433 mole of calcium nitrate.

71.1 g Ca(NO3)2

Page 77: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 6.33 mol of NaHCO3

Page 78: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 6.33 mol of NaHCO3

532 g

Page 79: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 3.0 x 10-5 mole of sulfuric acid?

Page 80: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

What is the mass in grams of 3.0 x 10-5 mole of sulfuric acid?

2.9 x 10-3 g

Page 81: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many glucose molecules are in 5.23 g of C6H12O6

Page 82: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many glucose molecules are in 5.23 g of C6H12O6

1.75 x 10 22 molecules

Page 83: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atoms are in this sample?

Page 84: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many oxygen atoms are in this sample?

1.05 x 10 23 atoms O

Page 85: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Molecular Formula from Empirical Formula

If we know the molecular weight we can determine the molecular formula.

Page 86: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

28.) How do the substances in the molecular formula compare to the empirical formula.

The molecular formula subscripts are always a whole number multiple of the empirical formula subscripts.

Page 87: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene.

Page 88: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

C3H4 = 36 + 4 = 40

x 3

120

C9H12

Page 89: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Combustion Analysis

The empirical formula of a compound is based on experiments that give the number of moles of each element in a sample of the compound.

Page 90: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

29.) What is combustion analysis?

A method used to determine the empirical formula for compounds containing only hydrogen and carbon.

All of the carbon is converted to carbon dioxide and all of the hydrogen is converted to water.

Page 91: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

29.) What is combustion analysis?contd

The amount of CO2 and H2O produced is determined by mass increase.

If a third element is present, its mass can be determined by subtracting the masses of carbon and hydrogen from the original mass.

Page 92: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Isopropyl alcohol, a substance sold as rubbing alcohol is composed of C,H,and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol.

Page 93: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

C3H8O

Page 94: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C,H and O atoms. Combustion of a 0.225 g sample of this compound produces 0.512 g of CO2 and 0.209 g of H2O. A.) What is the empirical formula of caproic acid? B.) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

Page 95: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

A.) C3H6O

B.) C6H12O2

Page 96: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 6: Quantitative Information from Balanced Equations

The coefficients in a balanced chemical equation can be interpreted both as the relative number of molecules ( or formula units) involved in the reaction and as the relative numbers of moles.

The coefficients and the associated substances in a balanced equation are called stoichiometrically equivalent quantities.

Page 97: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 6: Quantitative Information from Balanced Equations

2H2 + O2→ 2H2O

2 mol H2≈ 1 mol O2≈ 2H2O

Page 98: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Because the substances in a balanced chemical equation are stoichiometrically equivalent, we can predict the amount of any other substance in the equation that will be needed or will be produced.

Page 99: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

30.) Predict the moles of H2O that will be produced from 1.57 mol of O2?

3.14 mole H2O

Page 100: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

31.) To predict the amount of any other substance that will be needed or will be produced follow these steps: 1.) Convert known value into moles. 2.) Convert from moles of your known to

moles of your unknown by using the coefficient in your balanced equation

3.) Convert from moles of unknown to the value requested.

Page 101: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

32.) Calculate the mass of CO2

produced when 1.00 gram of C4H10 is burned.

3.03 g CO2

Page 102: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

How many grams of water are produced in the oxidation of 1.00 gram of glucose, C6H12O6?

.6 g C6H12O6

Page 103: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory:2KClO3(s) → 2KCl (s) + 3O2(g)

How many grams of O2 can be prepared from 4.5 g of KClO3?

1.8 g O2

Page 104: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide to form solid carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide.

Page 105: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

.919 g carbon dioxide

Page 106: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g propane?

Page 107: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

3.64 g

Page 108: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Section 7: Limiting Reactants

In chemical reactions when one reactant is totally consumed the reaction stops.

Page 109: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

33.) What is the limiting reactant?

The reactant that is completely consumed in a reaction.

Page 110: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

34.) Which reactant determines the amount of product that can be formed ?

Limiting reactant

Page 111: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

35.) What do we call the other reactants?

The reactants in excess

Page 112: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

The most important commercial process for converting N2 from the air into nitrogen containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3) :How many mole of NH3 can be formed from 3.0 mole of N2 and 6.0 mole of H2

N2 + 3H2 → 2NH3

Page 113: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

4 mole of NH3

Page 114: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Consider the reaction 2Al + 3Cl2→ 2AlCl3

A mixture of 1.50 mole of Al and 3.00 mole of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) how many moles of AlCl3 are formed? ( c) How many moles of the excess reactant remain at the end of the reaction?

Page 115: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

a.) Al

b.) 1.5 mole of AlCl3

c.) .75 mole Cl2 remain.

Page 116: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Consider the following reaction that occurs in a fuel cell:

2 H2 + O2 → 2H2O

This reaction, done properly, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas. ( each measurement is given with two significant figures). How many grams of water can be formed?

Page 117: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

1400 grams of water.

Page 118: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

When working with limiting reactant problems, you will need to pick one product, determine the number of moles of that product that will be produced by each reactant. The reactant that gives the smallest amount number is the limiting reactant and determines the amount of product that can be formed.

Page 119: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

A strip of zinc metal with a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:

Zn + 2AgNO3 → 2Ag + Zn(NO3)2

Page 120: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Zn + 2AgNO3 → 2Ag + Zn(NO3)2

a.) Which reactant is the limiting reactant?

b) How many grams of Ag will form?

c.) How many grams of Zn(NO3)2 will form?

d) How many grams of the excess reactant will be left at the end of the reaction?

Page 121: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

Zn + 2AgNO3 → 2Ag + Zn(NO3)2

a.) AgNO3

b) 1.6 g Ag

c.) 1.39 g of Zn(NO3)3

d)1.53 g Zn left over.

Page 122: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

36. What is the theoretical yield?

The quantity of product that is calculated to form when all of the limiting reactant reacts.

Page 123: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

37.) What is the actual yield?

The amount of product that actually forms.

Page 124: Chemistry II Stoichiometry Chapter 3. Stoichiometery: Calculations with Chemical Formulas and Equations.

38.) What is the percent yield?

A comparison of the amounts of product that formed to the amount you expect to form.

% yield = Actual x 100

Theorectical