UNIT 6

Overview

 Reactions   Write formula/word equations Balance Equations  Identify Types   Predict Products Write Net Ionic Equations  Stoichiometry  Conversions   Limiting & Excess Reagent Percent Yield

Chemical Reactions



Process in which one or more pure substances are converted into one or more different pure substances Reactants: Zn + I 2 Product: Zn I 2

Indications of a Reaction

 Temperature Change  Color Change  Production of gas  Formation of a precipitate  Production of light

Chemical Equations

4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) (Reactants) (Products)



Reactants

react to produce

products

 The letters (s), (g), (l), and (aq) are the

physical states

of compounds.

 “aq” represents aqueous meaning dissolved in water (solution)  The numbers in the front are called

coefficients

.



Subscripts

compound represent the number of each atom in a

Chemical Reactions

Symbol

+ → ↔ ↑ ↓ ∆

Meaning

used to separate one reactant or product from another used to separate the reactants from the products - it is pronounced "yields" or "produces" when the equation is read used when the reaction can proceed in both directions - this is called an equilibrium arrow and will be used later in the course an alternative way of representing a substance in a gaseous state an alternative way of representing a substance in a solid state indicates that heat is applied to make the reaction proceed

Diatomic Elements

 Elements that cannot exist by themselves (always occur in pairs)  Bromine (Br 2 )  Iodine (I 2 )  Nitrogen (N 2 )  Chlorine (Cl 2 )  Hydrogen (H 2 )  Oxygen (O 2 )  Fluorine (F 2 )

Writing Equations Practice

1. When lithium hydroxide pellets are added to a solution of sulfuric acid, lithium sulfate and water are formed.

2 LiOH (s) + H 2 SO 4(aq)



Li 2 SO 4(aq) + 2 H 2 O (l)

2. When crystalline C formed.

6 H 12 O 6 is burned in oxygen, carbon dioxide and water vapor are

C 6 H 12 O 6(s) + 6 O 2(g)



6 CO 2(g) + 6 H 2 O (g)

Balancing Equations



Law of Conservation of Mass

 Matter cannot be destroyed (atoms of reactants must equal products)  Balance equations to get same number of each atom on the left and right in an equation

2HgO(s) ---> 2 Hg(l) + O

2

(g)

2 Hg atoms, 2 O atoms 2 Hg atoms, 2 O atoms

Balancing Equations

___ Al(s) + ___ Br 2 (l) ---> ___ Al 2 Br 6 (s) __C 3 H 8 (g) + __ O 2 (g)



__ CO 2 (g) + __ H 2 O(g) __ B 4 H 10 (g) + __ O 2 (g)



__ B 2 O 3 (g) + __ H 2 O(g)

6 Types of Reactions

 Synthesis (combination)  Decomposition  Single Replacement (displacement)  Double Replacement (precipitation)  Combustion  Acid-Base Neutralization

Synthesis (Combination) Reactions

 Two or more substances combine to form a new compound.

A + X



AX Synthesis of:

Binary compounds Metal carbonates Metal hydroxides Metal chlorates Oxyacids H 2 CaO + CO 2 CaO + H 2 O  KCl + O CO + O 2 2 2   H 2 O  + H 2 O  CaCO Ca(OH) H 2 CO 3 3 2

Decomposition Reactions

 A single compound breaks down into two or more simpler substances

AX



A + X Decomposition of:

Binary compounds Metal carbonates Metal hydroxides Metal chlorates Oxyacids H 2 O  CaCO 3 H 2  Ca(OH) 2 KClO 3  H 2 CO 3  + O 2 CaO + CO 2  CaO + H 2 O KCl + O CO 2 2 + H 2 O

Single Replacement (displacement) Reactions

 One element replaces another in a reaction  Metals replace metals  Nonmetals replace nonmetals

A + BX



BX + Y



AX + B BY + X

Activity Series

 Decide whether or not one element will replace another  Metals can replace other metals provided that they are above the metal that they are trying to replace  If the metal is not above what it is trying to replace, the result is “no reaction”

Double Replacement (Precipitation) Reactions

 Two elements or ions “switch partners”

AX + BY



AY + BX

 One of the compounds formed is usually a

precipitate

, an

insoluble gas

that bubbles out of solution, or a

molecular compound

, usually water.

Solubility



Solubility

– ability to dissolve  In a double replacement (precipitate) reaction, one of the products must be insoluble in water and form a precipitate 

Precipitate

solution – insoluble solid formed by a reaction in  If both products are soluble the result is “no reaction”  Solubility rules help you determine whether or not a compound will form a precipitate or remain an aqueous solution

Solubility Rules

Soluble Ionic Compounds

Alkali metals, NH 4 + NO 3 , C 2 H 3 O 2 , ClO 3 , ClO 4 Cl , Br , I SO 4 -2

Except with:

(no exceptions) Ag + , Hg 2 +2 , Pb +2 Sr +2 , Ba +2 , Ca +2 , Ag + , Pb +2 , Hg 2 +2

Insoluble Ionic Compounds

CO 3 -2 , PO 4 -3 , SiO 3 -2 , O -2 , SO 3 -2 , CrO 4 -2 S -2 OH -

Except with:

NH 4 + , alkali metals NH 4 + , alkali metals Ca +2 , Sr +2 , Ba +2 , Mg +2 (group 2) NH 4 + , alkali metals, (Ca +2 , Ba +2 , Sr +2 are slightly soluble)

Combustion Reactions

 A substance combines with oxygen, releasing a large amount of energy in the form of light and heat.

 Produces a flame  Fuel + oxygen produces carbon dioxide and water vapor

C x H x + O 2



CO 2 + H 2 O

Acid-Base Neutralization Reactions

 When the solution of an acid and solution of a base are mixed   Products have no characteristics of either the acid or the base Acid + Base (metal hydroxide)  salt + water  Salt comes from cation of base and anion of acid

HY + XOH



XY + H 2 O

Chemical Equations



Molecular Equation

– shows complete chemical formulas of reactants and products Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq) 

Complete Ionic Equation

– All soluble electrolytes shown as ions Pb +2 (aq) + 2NO 3 (aq) + 2K + (aq) + 2I (aq)  PbI 2 (s) + 2K + (aq) + 2NO 3 (aq) 

Net Ionic Equation

– shows only the ions and molecules directly involved in the equation Pb +2 (aq) + 2I (aq)  PbI 2 (s)

Writing Complete Ionic Equations

1.

2.

 Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions. indicate the correct formula and charge of each ion   indicate the correct number of each ion write (aq) after each ion 3.

Start with a balanced molecular equation. Bring down all compounds with (s), (l), or (g) unchanged.

Writing Complete Ionic Equations

Example: 2Na 3 PO 4 (aq) + 3CaCl 2 (aq)  6NaCl(aq) + Ca 3 (PO 4 ) 2 (s) Becomes… 6Na + (aq) + 2PO 4 3 (aq) + 3Ca 2+ (aq) + 6Cl (aq)  6Cl (aq) + Ca 3 (PO 4 ) 2 (s) 6Na + (aq) +

Spectator Ions

 Appear in identical forms among both the reactants and products of a complete ionic equation  When writing

net ionic equations

cancel each other out they Pb +2 (aq) + 2NO 3 (aq) + 2K + (aq) + 2I (aq)  PbI 2 (s) + 2K + (aq) + 2NO 3 (aq)

Writing Net Ionic Equations

 Cancel out spectator ions from complete ionic equation then write what’s left 6Na + (aq) + 2PO 4 3 (aq) + 3Ca 2+ (aq) + 6Cl (aq) 6Cl (aq) + Ca 3 (PO 4 ) 2 (s)  6Na + (aq) + Becomes… 2PO 4 3 (aq) + 3Ca 2+ (aq)  Ca 3 (PO 4 ) 2 (s)

Practice

Write complete ionic and net ionic equations for the following: 1.

3(NH 4 ) 2 CO 3 (aq) + 2Al(NO 3 ) 3 (aq)  Al 2 (CO 3 ) 3 (s) 6NH 4 NO 3 (aq) + 2.

2NaOH(aq) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + 2H 2 O(l) 3.

Zn(s) + CuSO 4 (aq) --> ZnSO 4 (aq) + Cu(s)

1.

Answers

Complete Ionic Equation: 6NH 4 + (aq) + 3CO 3 2 (aq) + 2Al 3+ (aq) + 6NO 3 (aq)  6NH 4 + (aq) + 6NO 3 (aq) + Al 2 (CO 3 ) 3 (s) Net Ionic Equation: 2 Al 3+ (aq) + 3 CO 3 2 (aq)  Al 2 (CO 3 ) 3 (s) 2.

Complete Ionic Equation: 2Na + (aq) + 2OH (aq) + 2H + (aq) + SO 4 2 (aq)  2Na + (aq) + SO 4 2 (aq) + 2H 2 O(l) Net Ionic Equation: OH (aq) + H + (aq)  H 2 O(l)

*Note: simplify net ionic equations if possible

3.

Complete Ionic Equation: Zn(s) + Cu 2+ (aq) + SO 4 2 (aq)  Zn 2+ (aq) + SO 4 2 (aq) + Cu(s) Net Ionic Equation: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Stoichiometry

The study of the quantitative aspects of chemical reactions.

Mole Ratio

 Conversion factor that relates amount in moles of any two substances involved in a chemical reaction 2 Al 2 O 3 (l)  4 Al(s) + 3 O 2 (g) Mole ratio Al 2 O 3 to O 2 = 2:3 Mole ratio Al to Al 2 O 3 Mole ratio Al to O 2 = 4:2 or 2:1 = 4:3

Stoichiometry Problems

 Solved just like conversions!

 You must start with a balanced chemical equation  Types:  Mole  Mole  Mass  Mass  Mass  Mole or Mole  Mass

Mole



Mole

2 Al 2 O 3 (l)  4 Al(s) + 3 O 2 (g)  How many moles of O 2 3.5 moles of Al 2 O 3 ?

are produced from 3.5 mol Al 2 O 3 × 2 3 mol O 2 mol Al 2 O 3 = 5.25 mol O 2 *

Use mole ratio to convert between moles!

Mass



Mass

2 Al 2 O 3 (l)  4 Al(s) + 3 O 2 (g)  How many grams of Al are produced from 4.56 grams of Al 2 O 3 ?

 Molar Mass Al 2 O 3 = 101.96 g/mol Molar Mass Al = 26.98 g/mol 4.56 g Al 2 O 3 × 1 mol Al 2 O 3 101.96 g Al 2 O 3 × 4 mol Al 2 mol Al 2 O 3 × 26.98 g Al 1 mol Al = 2.41 g Al

Limiting/Excess Reactant

Recipe makes 10 pancakes  3 eggs  2 cups bisquik  1 cup milk  1 cup chocolate chips What “limits” how many pancakes I can make and what will be left over?



What is the most amount of pancakes that I can make with 6 eggs and 5 cups of milk?



What is the most amount of pancakes that I can make with 3 cups of chocolate chips and 8 cups of milk?

Limiting/Excess Reactant

 The

limiting reactant

is the reactant that is

consumed first

, limiting the amounts of products formed.

 The

excess reactant

is the reactant that is leftover after the reaction has gone to completion.

Limiting/Excess Reactant

Reactants Products

2 NO(g) + O

2

(g) 2 NO

2

(g)

Limiting reactant = ___________ Excess reactant = ____________

Calculating Limiting/Excess Reagent

2 NO(g) + O

2

(g) 2 NO

2

(g)

 Given 12.4 grams of NO and 9.40 grams of O 2 , which is the limiting and which is the excess reagent?

12.4 g NO × 1 mol NO 30.01 g NO × 2 mol NO 2 2 mol NO × 46.01 g NO 2 1 mol NO 2 = 19.01 g NO 2 9.40 g O 2 × 1 mol O 2 16.00 g O 2 × 2 mol NO 2 1 mol O 2 × 46.01 g NO 2 1 mol NO 2 = 54.06 g NO 2

Calculating Limiting/Excess Reagent

2 NO(g) + O

2

(g) 2 NO

2

(g)

12.4 g NO × 1 mol NO 30.01 g NO × 2 mol NO 2 2 mol NO × 46.01 g NO 2 1 mol NO 2 = 19.01 g NO 2 9.40 g O 2 × 1 mol O 2 32.00 g O 2 × 2 mol NO 2 1 mol O 2 × 46.01 g NO 2 1 mol NO 2 = 27.03 g NO 2  NO

limits

the amount of NO 2  Limiting reagent = NO that is made  O 2 will be

leftover

once the reaction is complete  Excess reactant = O 2

Calculating Limiting/Excess Reagent

2 NO(g) + O

2

(g) 2 NO

2

(g)

 How much O 2 will be in excess once the reaction is complete?

12.4 g NO × 1 mol NO 30.01 g NO × 1 mol O 2 2 mol NO × 32.00 g O 2 1 mol O 2 = 6.61 g O 2 6.61 grams of O 2 will be

used

in the reaction. You have 9.40 grams to start with.

9.40 – 6.61 = 2.79 grams O 2 in excess (leftover)

Limiting/Excess Reactant



If the equation has 2 or more products, when determining the limiting/excess reactants, simply pick one of the products and convert both reactants to that product.

 You MUST use the same product for both.

Percent Yield

Actual Yield Theoretical Yield × 100  Percentage comparing how much product was

actually

produced compared to what

should have been

produced.

 Calculate theoretical yield using stoichiometry.  If you know how much of each reactant you start out with, use stoichiometry to calculate how much of the given product you should produce.

Percent Yield

AgNO 3 (aq) + KCl (aq) → AgCl(s) + KNO 3 (aq) An experiment was performed combining using 3.4 g of AgNO 3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment?

3.4 g AgNO 3 × 1 mol AgNO 3 169.88 g AgNO 3 × 1 mol AgCl 1 mol AgNO 3 × 143.32 g AgCl 1 mol AgCl = 2.9 g AgCl 2.7

2.9

× 100 = 93%