# Reactions & Stoichiometry

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Reactions & Stoichiometry
UNIT 6 Reactions & Stoichiometry

Overview Reactions Stoichiometry 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 + I2 Product: Zn I2

Indications of a Reaction
Temperature Change Color Change Production of gas Formation of a precipitate Production of light

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
Chemical Equations 4 Al(s) O2(g) ---> 2 Al2O3(s) 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 represent the number of each atom in a compound (Reactants) (Products)

Chemical Reactions + → ↔ ↑ ↓ ∆ Meaning
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

BrINClHOF! Diatomic Elements
Elements that cannot exist by themselves (always occur in pairs) Bromine (Br2) Iodine (I2) Nitrogen (N2) Chlorine (Cl2) Hydrogen (H2) Oxygen (O2) Fluorine (F2) BrINClHOF!

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) + H2SO4(aq)  Li2SO4(aq) + 2 H2O(l) 2. When crystalline C6H12O6 is burned in oxygen, carbon dioxide and water vapor are formed. C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)

Balancing Equations 2HgO(s) ---> 2 Hg(l) + O2(g)
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 2 Hg atoms, 2 O atoms Hg atoms, 2 O atoms 2HgO(s) ---> 2 Hg(l) + O2(g)

Balancing Equations ___ Al(s) + ___ Br2(l) ---> ___ Al2Br6(s) __C3H8(g) + __ O2(g)  __ CO2(g) + __ H2O(g) __ B4H10(g) + __ O2(g)  __ B2O3(g) + __ H2O(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 H2 + O2  H2O Metal carbonates CaO + CO2  CaCO3 Metal hydroxides CaO + H2O  Ca(OH)2 Metal chlorates KCl + O2  KClO3 Oxyacids CO2 + H2O  H2CO3

Decomposition Reactions
A single compound breaks down into two or more simpler substances AX  A + X Decomposition of: Binary compounds H2O  H2 + O2 Metal carbonates CaCO3  CaO + CO2 Metal hydroxides Ca(OH)2  CaO + H2O Metal chlorates KClO3  KCl + O2 Oxyacids H2CO3  CO2 + H2O

Single Replacement (displacement) Reactions
One element replaces another in a reaction Metals replace metals Nonmetals replace nonmetals A + BX  AX + B BX + Y  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 – insoluble solid formed by a reaction in solution 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 Except with:
Alkali metals, NH4+ NO3-, C2H3O2-, ClO3-, ClO4- (no exceptions) Cl-, Br-, I- Ag+, Hg2+2, Pb+2 SO4-2 Sr+2, Ba+2, Ca+2, Ag+, Pb+2, Hg2+2 Insoluble Ionic Compounds Except with: CO3-2, PO4-3, SiO3-2, O-2, SO3-2, CrO4-2 NH4+, alkali metals S-2 Ca+2, Sr+2, Ba+2, Mg+2 (group 2) OH- NH4+, alkali metals, (Ca+2, Ba+2, Sr+2 are slightly soluble)

Combustion Reactions CxHx + O2  CO2 + H2O
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 CxHx + O2  CO2 + H2O

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 + H2O

Chemical Equations Molecular Equation – shows complete chemical formulas of reactants and products Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq) Complete Ionic Equation – All soluble electrolytes shown as ions Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq) Net Ionic Equation – shows only the ions and molecules directly involved in the equation Pb+2(aq) + 2I-(aq)  PbI2(s)

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

Writing Complete Ionic Equations
Example: 2Na3PO4(aq) + 3CaCl2(aq)  6NaCl(aq) + Ca3(PO4)2(s) Becomes… 6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6Na+(aq) + 6Cl-(aq) + Ca3(PO4)2(s)

Spectator Ions Appear in identical forms among both the reactants and products of a complete ionic equation When writing net ionic equations they cancel each other out Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)

Writing Net Ionic Equations
Cancel out spectator ions from complete ionic equation then write what’s left 6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6Na+(aq) + 6Cl-(aq) + Ca3(PO4)2(s) Becomes… 2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)

Practice Write complete ionic and net ionic equations for the following: 3(NH4)2CO3(aq) + 2Al(NO3)3(aq)  6NH4NO3(aq) + Al2(CO3)3(s) 2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l) Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)

6NH4+(aq) + 3CO32-(aq) + 2Al3+(aq) + 6NO3-(aq)  6NH4+(aq) + 6NO3-(aq) + Al2(CO3)3(s) Net Ionic Equation: 2 Al3+(aq) + 3 CO32-(aq)  Al2(CO3)3(s) 2Na+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq)  2Na+(aq) + SO42-(aq) + 2H2O(l) Net Ionic Equation: OH-(aq) + H+(aq)  H2O(l) *Note: simplify net ionic equations if possible Zn(s) + Cu2+(aq) + SO42-(aq)  Zn2+(aq) + SO42-(aq) + Cu(s) Net Ionic Equation: Zn(s) + Cu2+(aq)  Zn2+(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 Al2O3(l)  4 Al(s) + 3 O2(g) Mole ratio Al2O3 to O2 = 2:3 Mole ratio Al to Al2O3 = 4:2 or 2:1 Mole ratio Al to O2 = 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 Al2O3(l)  4 Al(s) + 3 O2(g)
How many moles of O2 are produced from 3.5 moles of Al2O3? 3.5 mol Al2O3 × = 5.25 mol O2 *Use mole ratio to convert between moles! 3 mol O2 2 mol Al2O3

Mass  Mass 2 Al2O3(l)  4 Al(s) + 3 O2(g)
How many grams of Al are produced from 4.56 grams of Al2O3? Molar Mass Al2O3 = g/mol Molar Mass Al = g/mol 4.56 g Al2O3 × × × = 2.41 g Al 1 mol Al2O3 g Al2O3 4 mol Al 2 mol Al2O3 26.98 g Al 1 mol Al

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

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) + O2 (g) NO2(g) Limiting reactant = ___________ Excess reactant = ____________

Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g) NO2(g) Given 12.4 grams of NO and 9.40 grams of O2, which is the limiting and which is the excess reagent? 1 mol NO 30.01 g NO 2 mol NO2 2 mol NO 46.01 g NO2 1 mol NO2 12.4 g NO × × × = g NO2 1 mol O2 16.00 g O2 2 mol NO2 1 mol O2 46.01 g NO2 1 mol NO2 9.40 g O2 × × × = g NO2

Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g) NO2(g) 1 mol NO 30.01 g NO 2 mol NO2 2 mol NO 46.01 g NO2 1 mol NO2 12.4 g NO × × × = g NO2 1 mol O2 32.00 g O2 2 mol NO2 1 mol O2 46.01 g NO2 1 mol NO2 9.40 g O2 × × × = g NO2 NO limits the amount of NO2 that is made Limiting reagent = NO O2 will be leftover once the reaction is complete Excess reactant = O2

Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g) NO2(g) How much O2 will be in excess once the reaction is complete? 1 mol NO 30.01 g NO 1 mol O2 2 mol NO 32.00 g O2 1 mol O2 12.4 g NO × × × = 6.61 g O2 6.61 grams of O2 will be used in the reaction. You have 9.40 grams to start with. 9.40 – 6.61 = 2.79 grams O2 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 × 100 Theoretical Yield
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.

AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)
Percent Yield AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) An experiment was performed combining using 3.4 g of AgNO3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment? 1 mol AgNO3 g AgNO3 1 mol AgCl 1 mol AgNO3 g AgCl 1 mol AgCl 3.4 g AgNO3 × × × = 2.9 g AgCl 2.7 2.9 Percent Yield = × 100 = 93%