5. 1 Introduction to Chemical Reactions A 5.1 Introduction to Chemical Reactions A. General Features of Physical and Chemical Changes A chemical change (a chemical reaction) converts one substance into another. Chemical reactions involve: Breaking bonds in the reactants (starting materials) Forming new bonds in the products

5. 1 Introduction to Chemical Reactions A 5.1 Introduction to Chemical Reactions A. General Features of Physical and Chemical Changes

5.1 Introduction to Chemical Reactions B. Writing Chemical Equations A chemical equation uses chemical formulas and other symbols showing what reactants are the starting materials in a reaction and what products are formed.

5.1 Introduction to Chemical Reactions B. Writing Chemical Equations The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction.

5.1 Introduction to Chemical Reactions B. Writing Chemical Equations

5.2 Balancing Chemical Equations HOW TO Balance a Chemical Equation Write a balanced chemical equation for the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). Example Step [1] Write the equation with the correct formulas.

5.2 Balancing Chemical Equations HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Step [2] Balance the C’s first: Balance the H’s next:

5.2 Balancing Chemical Equations HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Step [2] Finally, balance the O’s:

Balance the Equations __H2 (g) + __O2  __H20 __NO + __O2  __NO2 __CH4 + __Cl2  __CH2Cl2 + __HCl

Polyatomic Ions __Ca3(PO4)2 + __H2SO4  __CaSO4 + __H3PO4

Balance the Equation __Al + __H2SO4  __Al2(SO4)3 + __H2 __Na2SO3 + __H3PO4  __H2SO3 + __Na3PO4 __Al + __H2SO4  __Al2(SO4)3 + __H2 __NaSO3 + __H3PO4  __H2SO3 + __Na3PO4

Balance the Equation __Mg + __HBr  __MgBr2 + __H2 __KClO3  __KCl + __O2 __CH4 + __Cl2  __CCl4 + __HCl __Mg + __HBr  __MgBr2 + __H2 __KClO3  __KCl + __O2 __CH4 + __Cl2 __CCl4 + __HCl

Balance the Equation __Al2O3 + __HCl  __AlCl3 + __H2O __Al(OH)3 + __H2SO4  __Al2(SO4)3 + __H2O

__Ni + __HCl  __NiCl2 + __H2 __PbS + __O2  __PbO + __SO2 __H3PO4 + __Ca(OH)2  __Ca3(PO4)2 + __H2O

__H2SO4 + __NaOH  __Na2SO4 + __H2O __CO + __O2  __CO2 __S + __O2 + __H2O  __H2SO4

5.3 Types of Reactions The majority of chemical reactions fall into 6 categories: combination decomposition single replacement double replacement oxidation and reduction (Section 5.4) acid-base (Chapter 9)

5.3 Types of Reactions A. Combination and Decomposition A combination reaction is the joining of two or more reactants to form a single product.

5.3 Types of Reactions A. Combination and Decomposition A decomposition reaction is the conversion of a single reactant to two or more products.

5.3 Types of Reactions B. Replacement Reactions A single replacement reaction is a reaction in which one element replaces another element in a compound to form a different compound and element as products.

5.3 Types of Reactions B. Replacement Reactions

5.3 Types of Reactions B. Replacement Reactions A double replacement reaction is a reaction in which two compounds exchange “parts”–atoms or ions—to form two new compounds.

5.3 Types of Reactions B. Replacement Reactions

Combination, Decomposition, Single Displacement or Double Displacement? Ni(NO3)2 + Mg  Ni + Mg(NO3)2 2 KI + Sn(NO3)2  SnI2 + 2 KNO3 2 HgO  2 Hg + O2

Predicting Reactions

Predicting Reactions Combination: N2 + ____  Mg3N2 Decomposition: 2 SO3  2SO2 + ______ Single Replacement: 2 Ag + CuBr2  ______ + ______ Double Replacement: KOH + HI  ______ + _______

5.4 Oxidation and Reduction A. General Features Oxidation is the loss of electrons from an atom. Reduction is the gain of electrons by an atom. Both processes occur together in a single reaction called an oxidation−reduction or redox reaction. A redox reaction involves the transfer of electrons from one element to another. A redox reaction always has two components, one that is oxidized and one that is reduced.

5.4 Oxidation and Reduction A. General Features

5.4 Oxidation and Reduction A. General Features Zn + Cu2+ Zn2+ + Cu Each of these processes can be written as an individual half reaction: Oxidation half reaction: Reduction half reaction:

5.4 Oxidation and Reduction A. General Features Zn + Cu2+ Zn2+ + Cu oxidized reduced A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent. Cu2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized.

5.4 Oxidation and Reduction A. General Features Zn + Cu2+ Zn2+ + Cu oxidized reduced A compound that is oxidized while causing another compound to be reduced is called a reducing agent. Zn acts as a reducing agent because it causes Cu2+ to gain electrons and become reduced.

5.4 Oxidation and Reduction A. General Features

5. 4 Oxidation and Reduction B 5.4 Oxidation and Reduction B. Examples of Oxidation–Reduction Reactions Iron Rusting O gains e– and is reduced. 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) neutral Fe neutral O Fe3+ O2– Fe loses e– and is oxidized.

5. 4 Oxidation and Reduction B 5.4 Oxidation and Reduction B. Examples of Oxidation–Reduction Reactions Zn + 2 MnO2 ZnO + Mn2O3

Zn + 2H+  Zn2+ + H2 Fe3+ + Al  Al3+ + Fe

I- + Br2  I2 + Br- AgBr  Ag + Br2

5. 4 Oxidation and Reduction B 5.4 Oxidation and Reduction B. Examples of Oxidation–Reduction Reactions Oxidation results in the: Reduction results in the: Gain of oxygen atoms Loss of hydrogen atoms Loss of oxygen atoms Gain of hydrogen atoms

5.5 The Mole and Avogadro’s Number A mole is a quantity that contains 6.02 x 1023 items. 1 mole of C atoms = 6.02 x 1023 C atoms 1 mole of H2O molecules = 6.02 x 1023 H2O molecules 1 mole of Vitamin C molecules = 6.02 x 1023 Vitamin C molecules The number 6.02 x 1023 is Avogadro’s number. 1.3×1023 kg Titan, largest moon of Saturn 1.5×1023 kg Ganymede, largest moon of Jupiter 3.3×1023 kg Mercury 6.4×1023 kg Mars

How many items do 1 mol of the following contain: Baseballs Bicycles Cheerios CH4 molecules

5.5 The Mole and Avogadro’s Number It can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules: 1 mol 6.02 x 1023 atoms or 6.02 x 1023 atoms 1 mol 1 mol 6.02 x 1023 molecules or 6.02 x 1023 molecules 1 mol

5.5 The Mole and Avogadro’s Number Sample Problem 5.5 How many molecules are contained in 5.0 moles of carbon dioxide (CO2)? Identify the original quantity and the desired quantity. Step [1]

5.5 The Mole and Avogadro’s Number Step [2] Write out the conversion factors. Step [3] Set up and solve the problem.

How many C atoms are there in the following: 2.0 mol 6.0 mol 0.5 mol 25.0 mol

How many molecules are contained in each of the following number of moles 2.5mol of penicillin 0.25 mol of NH3 0.4 mol of Sugar 55.3 mol of Acetaminophen

5.6 Mass to Mole Conversions The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units (amu). HOW TO Calculate the Formula Weight of a Compound Example Calculate the formula weight for FeSO4. Step [1] Write the correct formula and determine the number of atoms of each element from the subscripts.

5.6 Mass to Mole Conversions HOW TO Calculate the Formula Weight of a Compound Multiply the number of atoms of each element by the atomic weight and add the results. Step [2]

5.6 Mass to Mole Conversions A. Molar Mass The molar mass is the mass of one mole of any substance, reported in grams per mole (g/mol). The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

5.6 Mass to Mole Conversions B. Relating Grams to Moles The molar mass relates the number of moles to the number of grams of a substance. In this way, molar mass can be used as a conversion factor. The molar mass of H2O is 18.0 g/mol, the conversion factor can be written:

5.6 Mass to Mole Conversions B. Relating Grams to Moles Sample Problem 5.9 How many moles are present in 100. g of aspirin (C9H8O4)? Step [1] Calculate the molar mass.

5.6 Mass to Mole Conversions B. Relating Grams to Moles Step [2] Write out the conversion factors. The conversion factor is the molar mass, and it can be written in two ways. Choose the one that places the unwanted unit, grams, in the denominator so that the units cancel:

5.6 Mass to Mole Conversions B. Relating Grams to Moles Step [3] Set up and solve the problem.

How many moles are contained in the following: 100 g NaCl

How many moles are contained in the following: 0.25g Aspirin(C9H8O4)

How many moles are contained in the following: 25.5g CH4

How many moles are contained in the following: 25g of H2O

5. 6 Mass to Mole Conversions C 5.6 Mass to Mole Conversions C. Relating Grams to Number of Atoms or Molecules We can also use the molar mass to show the relationship between grams and number of molecules (or atoms).

5. 6 Mass to Mole Conversions C 5.6 Mass to Mole Conversions C. Relating Grams to Number of Atoms or Molecules Sample Problem 5.10 How many molecules are in a 325-mg tablet of aspirin (C9H8O4)? Step [1] Find the molar mass

5. 6 Mass to Mole Conversions C 5.6 Mass to Mole Conversions C. Relating Grams to Number of Atoms or Molecules Step [2] Write out the conversion factors.

5. 6 Mass to Mole Conversions C 5.6 Mass to Mole Conversions C. Relating Grams to Number of Atoms or Molecules Step [3] Set up and solve the problem.

How many molecules are present in a 500mg tablet of penicillin (C16H18N2O4S)

How many molecules are present in a 500mg tablet of penicillin (C16H18N2O4S)

How many molecules are present in a 750mg of Mescaline (Hallucinogenic from peyote) (C11H17NO3)

5.7 Mole Calculations in Chemical Equations A balanced chemical equation also tell us: The number of moles of each reactant that combine The number of moles of each product formed 1 N2(g) + 1 O2(g) 2 NO(g) (The coefficient “1” has been written for emphasis.)

5.7 Mole Calculations in Chemical Equations Coefficients are used to form mole ratios, which can serve as conversion factors. N2(g) + O2(g) 2 NO(g) Mole ratios:

5.7 Mole Calculations in Chemical Equations Sample Problem 5.11 Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C2H6? 2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) Step [1] Identify the original and desired quantities.

5.7 Mole Calculations in Chemical Equations 2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) Step [2] Write out the conversion factors. Step [3] Set up and solve the problem.

Complete the following conversions + O2(g) 2 NO(g) How many mol of NO are formed from 3.3 mol of N2 How many mol of NO are formed from 0.5mol of O2 How many moles of O2 are needed to completely react with 1.2 mol of N2

N2(g) + O2(g) 2 NO(g) How many mol of NO are formed from 3.3 mol of N2 How many mol of NO are formed from 0.5mol of O2

N2(g) + O2(g) 2 NO(g) How many moles of O2 are needed to completely react with 1.2 mol of N2

5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Using the balanced equation, how many grams of O3 are formed from 9.0 mol of O2. Example sunlight 3 O2(g) 2 O3(g)

5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor. Step [1] sunlight 3 O2(g) 2 O3(g)

5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Convert the number of moles of product to the number of grams of product using the product’s molar mass. Step [2]

5.8 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Set up and solve the conversion. sunlight 3 O2(g) 2 O3(g)

C6H12O6 2 C2H6O + 2CO2 (glucose) (ethanol) How many g of ethanol are formed from 0.55 mol of glucose How many g of CO2 are formed from 0.25 mol of glucose How many g of glucose are needed to form 1mol of ethanol

How many g of ethanol are formed from 0.55 mol of glucose C6H12O6 2 C2H6O + 2CO2

How many g of CO2 are formed from 0.25 mol of glucose C6H12O6 2 C2H6O + 2CO2

How many mol of ethanol would be formed if 5g of glucose is used C6H12O6 2 C2H6O + 2CO2

5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Ethanol (C2H6O) is synthesized by reacting ethylene (C2H4) with water. How many grams of ethanol are formed from 14 g of ethylene? Example C2H4 + H2O C2H6O

5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product C2H4 + H2O C2H6O

5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product C2H4 + H2O C2H6O

5.8 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product C2H4 + H2O C2H6O

2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) 1) How many molecules of CO is produced? 2) How many g of C2H6 are needed to react with all of O2 3) How many g of H2O are produced when 15g of O2 is used.

4 CO(g) + 6 H2O(g) 2 C2H6(g) + 5 O2(g) How many molecules of CO is produced?

4 CO(g) + 6 H2O(g) 2 C2H6(g) + 5 O2(g) How many g of C2H6 are needed to react with all of O2

4 CO(g) + 6 H2O(g) 2 C2H6(g) + 5 O2(g) How many g of H2O are produced when 15g of O2 is used.

5.9 Percent Yield The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation. Usually, however, the amount of product formed is less than the maximum amount of product predicted. The percent yield is the amount of product isolated from a reaction.

5.9 Percent Yield Sample Problem 5.14 If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?

5.10 Limiting Reactants The limiting reactant is the reactant that is completely used up in a reaction.

5.10 Limiting Reactant A. Determining the Limiting Reactant Analyze the two possible outcomes: If the amount present of the second reactant is less than what is needed, the second reactant is the limiting reagent. If the amount present of the second reactant is greater than what is needed, the second reactant is in excess.

5.10 Limiting Reactant C. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 Using the balanced equation, determine the limiting reactant when 10.0 g of N2 (MM = 28.02 g/mol) react with 10.0 g of O2 (MM = 32.00 g/mol). N2(g) + O2(g) 2 NO(g)

5.10 Limiting Reactant C. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [1] Convert the number of grams of each reactant into moles using the molar masses.

5.10 Limiting Reactant C. Determining the Limiting Reactant Using the Number of Grams Sample Problem 5.20 [2] Determine the limiting reactant by choosing N2 as the original quantity and converting to mol O2. mole–mole Conversion factor 1 mol O2 1 mol N2 0.357 mol N2 x = 0.357 mol O2 The amount of O2 we started with (0.313 mol) is less than the amount we would need (0.357 mol) so O2 is the limiting reagent.

N2(g) + O2(g)  NO(g) Determine the limiting reactant under the following conditions: 1) 12.5g N2 and 15.0g O2 2) 14.0g N2 and 13.0g O2

N2(g) + O2(g)  NO(g) 12.5g N2 and 15.0g O2

N2(g) + O2(g)  NO(g) 14.0g N2 and 13.0g O2

Balance/Rxn Type/ REDOX Ni + HCl  NiCl2 + H2 CH4 + Cl2  CCl4 + HCl KClO3  KCl + O2 Al2O3 + HCl  AlCl3 + H2O

Given the following reaction: C12H22O11 + H2O  C2H6O + CO2 (sucrose) (ethanol) Balance the equation Determine the molecular weight of sucrose How many mols of Ethanol would be produced from 2 mols of sucrose? How many g of ethanol would be produced from 17.1 g sucrose?