Chapters+11+and+12

__** ~Chapters 11 and 12~ **__ Discovering Chemical Reactions and Stoichiometry

** ~Introduction~ **

=
This page has been constructed on the eleventh and twelfth chapters of Prentice Hall’s Chemistry textbook. Chapter eleven focuses on describing and identifying chemical reactions, including aqueous solutions. In this chapter, the process of writing skeleton equations is taught, and the five general types of chemical reactions are explored. Chapter eleven also deals with the reactions that occur in aqueous solutions, as well as predicting the formations of precipitates. Chapter twelve focuses on the calculations behind chemical equations. In this chapter, the arithmetic of equations is explored, including how balanced chemical equations can be used to calculate reactants and products of a reaction. Also explained in chapter twelve are mole ratios, actual yield, theoretical yield, percent yield, excess reagent, and limiting reagent. This wiki-page outlines these chapters in their entirety and includes information crucial to the understanding of these chapters. =====

** ~Groups~ **

 * __Editor:__ ** Eileen Corkery

__** Group 1: **__ Christos Anastos- pgs. 321-323 Lizzie Sieber- pgs. 324-326 Adam Shanahan (co-editor) pgs. 327-328

__** Group 2 **__ Marion Burdick- pgs. 330-332 Evan Grandfield (co-editor) pgs. 333-336 Nick Brault- pgs.337-339

__** Group 3 **__ Nina DeMeo (co-editor) pg.342 Tom DeMarco- pg. 343 Kendyl Barron- pg.344

__** Group 4 **__ Heather Bowditch (co-editor)- pictures Maggie Bie- pgs.353-355 Mike McShane- pgs.356-357

__** Group 5 **__ Courtney Gareau (co-editor)- pgs.359-362 Steven Denison- pgs.363-367

Katherine Perry- pgs.368-371 Andrew Sciotti- pgs. 372-375
 * __ Group 6 __**

~Material~ __** Group 1 (pgs.321-329) **__
 * By Christos Anastos, Lizzie Sieber, and Adam Shanahan **

[| http://richardbowles.tripod.com/chemistry/balance.htm] BALANCING EQUATION HELP

**__ Christos Anastos-Pgs 321-323 __** Writing Chemical Equations __ Chemical Reaction- __  a change in which one or more reactants change into one or more products; characterized by the breaking of bonds in reactants and the formation of bonds in products  -Chemical reactions are happening all the time  - Chemical equations help to convey as much information as possible about what  happens during a chemical reaction  Word Equations  - to write a word equation write the names of the reactants on the left side of the arrow, with different compounds or elements separated by plus sings. Write the names of the products on the right side of the arrow, with different compounds or elements separated by plus signs.  Chemical Equations  - Chemical Equations are written the same as word equations, except that instead  writing out the whole word for the reactant or product, you just right out it’s  chemical reaction.  Symbols used in Chemical Equations  - + - used to separate two reactants or two products  -an arrow- is used to represent a yield. A yield separates reactants from products.  - (s) - designates a reactant or product in the solid state; placed after the formula <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: normal; margin: 0in 0in 0pt;"> - (l) - designates a reactant or product in the liquid state; placed after the formula <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: normal; margin: 0in 0in 0pt;"> - (g) - designates a reactant or product in the gaseous state; placed after the formula <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: normal; margin: 0in 0in 0pt;"> - (aq) - designates an aqueous solution; placed after the formula <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: normal; margin: 0in 0in 0pt;"> - a triangle and an arrow- shows that heat was used in a reaction <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 115%; margin: 0in 0in 10pt;"> - chemical formula and an arrow- shows that the formula was used as a catalyst The Word Equation for rusting is: Iron+Oxygen -> Iron (III) Oxide

**__Elizabeth Sieber__** **__Pg. 324-326__** Balancing Chemical Equations · Unbalanced equations do not show the quantities you need of each o Ex. F+W+H+P à FW2HP2 · A balanced equation shows you the quantity (or moles) of each element o Some number of atoms on each side o Mass is conserved o Ex. F+2W+H+2P à FW2HP2 · Coefficients are the small whole numbers placed before the molecules. o Ex. The 2 in 2W o Or The 2 in 2P Video: media type="youtube" key="RnGu3xO2h74" height="390" width="640"

__**﻿Pages 327-328 - Adam Shanahan**__

Rules for Writing and Balancing Equations

 * 1) Determine the correct formulas for all reactants and products.
 * 2) Write the skeleton equation by placing the formulas for the reactants on the left, and the formulas for the products on the right. Place a yield sign (-->) in between them.
 * 3) Determine the # of atoms of each element in the reactants and products. Count polyatomic ions as single units if they are unchanged on both sides of the equation.
 * 4) Balance the elements by using //coefficients// (the numbers to the left of the element) and not //subscripts// (the small numbers near the bottom right of the element)
 * 5) Check to make sure that that each atom/polyatomic ion are equal on both sides of the equation,
 * 6) Reduce the coefficients to the lowest possible ratio.

__Example__
Hydrogen and oxygen react to form water. The reaction releases enough energy to launch a rocket. Write a balanced equation for the reaction.
 * 1) The only part of the sentence that we need to use is the first one. It says Hydrogen (H) and Oxygen (O) react to form water (H ﻿ ﻿2 O). Since Hydrogen and Oxygen create the water, they are the reactants and thus are on the right side. Since water is what is created it is the product and thus is on the right side. We also know that Oxygen and Hydrogen are diatomic (when they are alone they end in subscript 2) so the formula for Hydrogen is H ﻿2 the formula for oxygen is O ﻿﻿ ﻿2  and the formula for hydrogen is H ﻿2 O
 * 2) So this is the easy part, we just put it all together. Not really sure how to explain this.... but I figure it is pretty obvious. H ﻿2 + O ﻿﻿ ﻿2  ---> H ﻿2 O
 * 3) Hydrogen has 2 atoms on both sides, but oxygen has 2 on the left side but only 1 on the right side. this means hydrogen is currently balanced but Oxygen is not.
 * 4) H ﻿2 + O ﻿﻿ ﻿2  ---> H ﻿2 O. To balance this equation we can only use //coefficients// not //subscripts// . We only need to fix Oxygen, and since Oxygen is lower on the right side, we have to add a //coefficient// to it. 1x2=2. so we make it H ﻿2 + O ﻿﻿ ﻿2  ---> 2H ﻿2 O [Obviously that would be FAR too easy, so of course the other elements of the equation get messed up every time you fix one]. Now we Oxygen balanced but Hydrogen has 4 on the right side and only 2 on the left. Using our 10 years of experience with math, we can expertly calculate that 2x2=4. Using this new found knowledge, we add a 2 to the Hydrogen of the left side. This makes the equation 2H ﻿2 + O ﻿﻿ ﻿2  ---> 2H ﻿2 O.
 * 5) there are 4 Hydrogens on both the left and right side, and 2 Oxygens on both the left and right side. This means the equation is balanced.
 * 6) Cannot reduce.

__** Group 2 (pgs.330-339) **__
 * By Marion Burdick, Evan Grandfield, and Nick Brault **

__** 11. 2 Types of Chemical Reactions **__ By: Marion Burdick

** Classifying Reactions ** **-** Sometimes reactions don’t fit just one of these categories - By recognizing a particular reaction you can: 1. See patterns of chemical behavior 2. Predict the products of reactions based off these patterns **<span style="-moz-background-clip: border; -moz-background-inline-policy: continuous; -moz-background-origin: padding; background: yellow none repeat scroll 0% 0%; font-family: Times New Roman;"> HELP!!! ** [| http://www.cat.pinellas.k12.fl.us/textbooks/chemistry/ebook/products/0-13-190443-4/chemasap/dswmedia/rsc/asap1_chem05_cmsm1112.htm]
 * ** The five general types of reaction are combination, decomposition, single-replacement, double-replacement, and combustion. **

** Combination Reactions ** //Definition:// A chemical change in which two or more substances react to form a single new substance. Ex. //CaO + H2O// à //Ca(OH)2//

Ex.

// General Rules: // - In these types of reactions the product is always a single substance which is a compound - Reactants: Metal + Nonmetal à Products: Compound of a metal cation with a nonmetal anion - Reactants: 2 Nonmetals à Products: 1 + possible products - Reactants: Transition metal + nonmetal à 1 + possible products **<span style="-moz-background-clip: border; -moz-background-inline-policy: continuous; -moz-background-origin: padding; background: yellow none repeat scroll 0% 0%; font-family: Times New Roman;"> HELP!!! ** [| http://www.cat.pinellas.k12.fl.us/textbooks/chemistry/ebook/products/0-13-190443-4/chemasap/dswmedia/rsc/asap1_chem05_cmcp1114.htm]

** Decomposition Reactions ** //Definition:// A chemical change in which a single compound breaks down into two or more simpler products. Ex. 2HgO (s) → 2Hg (l) + O2 (g) Ex.

// General Rules: // -Only **one** reactant à **2+** products -Usually requires energy to break down (heat, light, or electricity) **<span style="-moz-background-clip: border; -moz-background-inline-policy: continuous; -moz-background-origin: padding; background: yellow none repeat scroll 0% 0%; font-family: Times New Roman;"> HELP!!! ** [| http://www.cat.pinellas.k12.fl.us/textbooks/chemistry/ebook/products/0-13-190443-4/chemasap/dswmedia/rsc/asap1_chem05_cmcp1115.htm]

** Single-Replacement Reactions ** Definition: A chemical change in which one element replaces a second element in a compound. Ex. Cl2 + NaBr --> NaCl + Br2 In the above example bromine gets replaced/displaced by chlorine J

Ex.



//General rules:// - both sides of the reaction consist of an element and a product -whether on metal displaces another from a compound depends of the “relative reactivities”

** Activity Series ** Definition: A list of elements in order of decreasing activity (applies to metals/ halogens plus select nonmetals); the activity series of halogens is Fl, Cl, Br, I. Elements can **not** replace anything above them! In that situation a reaction does **not** occur. **<span style="-moz-background-clip: border; -moz-background-inline-policy: continuous; -moz-background-origin: padding; background: yellow none repeat scroll 0% 0%; font-family: Times New Roman;"> HELP!!! ** [| http://www.cat.pinellas.k12.fl.us/textbooks/chemistry/ebook/products/0-13-190443-4/chemasap/dswmedia/rsc/asap1_chem05_cmcp1117.htm]

__ Evan Grandfield __ __ Pgs. 333-336(including conceptual problem 118) __

__** Vocab **__
 * 1. single-replacement reaction-a chemical change in which one element replaces a second element in a compound; also called a displacement reaction **
 * 2. activity series-a list of elements in order of decreasing activity; the activity series of halogens is Fl, Cl, Br, I **
 * 3. double-replacement reaction-a chemical change that involves an exchange of positive ions between 2 compounds **
 * 4. combustion reaction-a chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light **


 * I. Single-Replacement Reactions **
 * A. A single-replacement reaction is a chemical change in which one element replaces a second element in a compound. **
 * 1. In a single-replacement reaction, both the reactants and the products consist of an element and a compound. **
 * 2. Ex. **


 * Zn(s) + Cu(NO3)2(aq) à Cu(s) + Zn(NO3)2(aq) **


 * a. In this single-replacement reaction, zinc and copper change places. **
 * b. Zn, the reacting element, replaces copper in the reactant compound Cu(NO3)2. **
 * c. Products are element Cu and compound Zn(NO3)2. **
 * B. The relative reactivities of the 2 metals determine whether one metal will displace another metal from a compound. **
 * 1. The activity series of metals lists metals in order of decreasing reactivity. **
 * a. A reactive metal will replace any metal with a lower reactivity. **
 * 2. Ex. Fe will displace Cu from a Cu compound in solution. **
 * a. Fe does NOT displace zinc. **
 * Activity series of metals **

<span style="display: block; height: 120px; line-height: normal; margin-bottom: 0in; margin-left: 45px; margin-top: 8px; position: absolute; width: 16px; z-index: 251659264;"> ||
 * || ** Name ** || ** Symbol ** ||
 * ** Decreasing activity ** || ** Lithium ** || ** Li ** ||
 * || ** Potassium ** || ** K **
 * || ** Calcium ** || ** Ca ** ||
 * || ** Sodium ** || ** Na ** ||
 * || ** Magnesium ** || ** Mg ** ||
 * || ** Aluminum ** || ** Al ** ||
 * || ** Zinc ** || ** Zn ** ||
 * || ** Iron ** || ** Fe ** ||
 * || ** Lead ** || ** Pb ** ||
 * || ** (Hydrogen) ** || ** (H) ** ||
 * || ** Copper ** || ** Cu ** ||
 * || ** Mercury ** || ** Hg ** ||
 * || ** Silver ** || ** Ag ** ||


 * b. Note on hydrogen: metals from Li to Na will replace H from acids and water; from Mg to Pb they will replace H from acids only **


 * C. A halogen can replace another halogen from a compound. **
 * 1. activity of halogens decreases as you go down Group 7A: F,Cl, Br, and I (in decreasing order). **

Conceptual Problem Writing Equations for Single-Replacement Reactions


 * -write a balanced chemical equation for each single-replacement reaction. **
 * Reactions that take place in aqueous solution: **
 * a. Zn(s) + H2SO4(aq) à **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> b. Cl2(aq) + NaBr(aq) à **


 * 1. Analyze **
 * a. According to the activity series of metals, Zn displaces H from an acid and takes its place. Balance the equation and remember H is diatomic. **
 * b. Cl more reactive than Br and displaces Br from its compounds. Balance the equation and remember Br is diatomic. **
 * 2. Solve **
 * -Write skeleton equation first…then apply rules for balancing equations **
 * a. Zn(s) + H2SO4(aq) à ZnSO4(aq) + H2(g) **
 * this is balanced **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> b. Cl2(aq) + NaBr(aq) à <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> NaCl(aq) + Br2(aq) **
 * need to balance it **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> Cl2(aq) + 2NaBr(aq) à <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 2NaCl(aq) + Br2(aq) **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: normal; margin-bottom: 0in;"> Balanced **

II. Double-Replacement Reactions
 * A. It happens once in a while that when 2 solutions of ionic compounds mix, nothing at all happens. However, at other times, the ions in the 2 solutions react. **
 * 1. A double-replacement reaction is a chemical change involving an exchange of positive ions between 2 compounds. **
 * a. Double-replacement reaction is sometimes referred to as double-displacement reaction. **
 * 2. Double-replacement reactions generally take place in aqueous solution and usually produce a precipitate, a gas, or a molecular compound like H2O. **
 * 3. One of the following (abc) must be true for a double-replacement reaction to occur. **
 * a. 1 of the products is slightly soluble and precipitates from solution. **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 1. Na2S(aq) + Cd(NO3)2(aq) à <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> CdS(s) + 2NaNO3(aq) **
 * a. This reaction of these aqeous solutions produces yellow precipitate of cadmium sulfide. **
 * b. 1 of the products is a gas. **
 * c. 1 product is a molecular compound like H2O. **

Conceptual Problem Writing Equations for Double-Replacement Reactions
 * -write a balanced chemical equation for each double-replacement reaction. **
 * a. CaBr2(aq) + AgNO3(aq) à (a precipitate of silver bromide is formed.) **
 * b. FeS(s) + HCl(aq) à (hydrogen sulfide gas (H2S) is formed). **


 * 1. Analyze **
 * a. force behind reaction = formation of precipitate. Must write correct formulas of the products using ionic charges and balance equation. **
 * b. A gas is formed so must use ionic charges to write correct formula of other product and balance the equation. **


 * 2. Solve **
 * -write skeleton and apply rules for balancing equations **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> a. CaBr2(aq) + AgNO3(aq) à <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> AgBr(s) + Ca(NO3)2(aq) **


 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> CaBr2(aq) + 2AgNO3(aq) à <span style="font-family: 'Times New Roman',serif; font-size: 12pt;"> 2AgBr(s) + Ca(NO3)2(aq) **
 * <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: normal; margin-bottom: 0in;"> Balanced! **


 * b. FeS(s) + HCl(aq) à H2S(g) + FeCl2(aq) **
 * FeS(s) + 2HCl(aq) à H2S(g) + FeCl2(aq) **
 * Balanced! **


 * Evan Grandfield- [| http://www.google.com/imgres?imgurl=http://blog.dearbornschools.org/harrisr2/files/2011/02/balanced-equation.gif&imgrefurl=http://blog.dearbornschools.org/harrisr2/&usg=__fx7Dwke-s5rVUEtxNe8mijzSs5Q=&h=266&w=363&sz=5&hl=en&start=0&sig2=UBXbvQvHz4PA7XVLcnBoEg&zoom=1&tbnid=H0JdbPwD1hFpPM:&tbnh=155&tbnw=212&ei=bfyDTZQRkpyBB6XflL0I&prev=/images%3Fq%3Dstudents%2Bbalancing%2Bchemical%2Bequations%26um%3D1%26hl%3Den%26biw%3D1366%26bih%3D653%26tbs%3Disch:1&um=1&itbs=1&iact=rc&oei=bfyDTZQRkpyBB6XflL0I&page=1&ndsp=18&ved=1t:429,r:7,s:0&tx=65&ty=83]**

III. Combustion Reactions
 * A. A combustion reaction is a chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light. They always involve oxygen as a reactant. **
 * 1. Usually the other reactant is hydrocarbon. **
 * a. Hydrocarbon is a compound composed of hydrogen and carbon. **
 * 2. Complete combustion of a hydrocarbon produces CO2­ and H2O. **
 * a. The combustion will not complete if the supply of oxygen is limited during a reaction. **
 * b. Carbon (soot) C and CO can also be products in addition to CO2­ and H2O. **
 * 3. Complete combustion of hydrocarbon releases a lot of energy as heat. **
 * a. Consequently, CH4 methane an and C3H8 propane are important fuels. **


 * B. Reactions between oxygen and some elements other than C are also examples of combustion reactions. **
 * 1. Magnesium and sulfur burn in oxygen’s presence and their reactions can be classified as combination reactions or combustion reactions. **


 * 2. Reactions: **
 * 2Mg(s) + O2(g) à 2MgO(s) **
 * S(s) + O2(g) à SO2(g) **


 * Conceptual Problem **
 * Writing Equations for Combustion Reactions **
 * -write balanced equations for the complete combustion of these compounds: **
 * a. benzene (C6H6(l)) b. Ethanol (CH3CH2OH(l)) **


 * 1. Analyze **
 * Oxygen=other reactant in combustion reactions **
 * CO2 and H2O=products **
 * -write skeleton equation and balance equation **


 * 2. Solve **
 * a. C6H6(l) + O2(g) à CO2(g) + H2O(g) **
 * not balanced! **


 * 2C6H6(l) + 15O2(g) à 12CO2(g) + 6H2O(g) **
 * Balanced! **


 * b. CH3CH2OH(l) + O2(g) à CO2(g) + H2O(g) **
 * not balanced! **


 * CH3CH2OH(l) + 3O2(g) à 2CO2(g) + 3H2O(g) **
 * Balanced! **

media type="youtube" key="Ph6xowy5fNc" height="390" width="480"
 * Evan Grandfield- video **


 * __ Group 3 (pgs.342-344) __**
 * By Nina DeMeo, Tom DeMarco, and Kendyl Barron **


 * Pg.342 - Nina DeMeo**
 * Net Ionic Equations **
 * __Key Concept:__ a net ionic equation shows only those particles involved in the reaction and is balanced with respect to both mass and charge**
 * - the world is water based **
 * > 70% of earth’s surface; 66% of an adult’s body; etc. **
 * - many chemical reactions also take place in water **
 * - in most ionic compounds, the reactant and one of the products disassociate into cations and anions when they dissolve in water **
 * - you can use the ions created to write a complete ionic equation, an equation that shows dissolved ionic compounds as disassociated free ions **

Net Ionic Equations
 * Thomas DeMarco, pg. 343 **

A spectator ion is an ion that is not directly involved in a chemical reaction. It appears on both sides, but is not changed. It is simple to remember by thinking of a spectator at a sporting event. The spectator is present but does not take part in any of the action. When a chemical reaction is written without the spectator ions, you get what is called the net ionic equation. The net ionic equation is the chemical reaction with the parts of the reaction that actually change or react. If the net ionic equation’s charges are not balanced, it is necessary to add coefficients to balance the charges.


 * Predicting the Formation of a Precipitate p.344 –Kendyl Barron**


 * -->> In a double replacement reaction mixing two ionic compounds can result in the formation of an insoluble salt known as a precipitate.**
 * -->>However, only some specific reactions produce precipitates. This depends on the solubility of the newly formed compound.**
 * -->>Follow the rules bellow to determine if a combination of ionic compounds will result in a precipitates. If the product of the reaction is insoluble, a precipitate will form.**


 * __ Solubility Rules for Ionic Compounds: __**
 * Salts of alkali metals and ammonia -->> soluble**
 * Nitrates salts and chlorate salts -->> soluble**
 * Most sulfate salts -->> soluble**
 * Most chloride salts -->> soluble**
 * Carbonates, phosphates, chromates,**
 * sulfides, and hydroxides -->> most insoluble.**




 * __ Group 4 (pgs.353-357) __**
 * By Heather Bowditch, Maggie Bie, and Mike McShane **

Maggie Bie – pg 353-354
__<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> Key concept: __ <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> **A balanced chemical equation provides the same kind of quantitative information that a recipe does**. <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> If you were going to make a tricycle: (example in book) <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> 1frame + 1 seat + 3 wheels + 1 handlebar + 2 pedals = Finished tricycle <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> F + S + 3W + 1H + 2P = FSW3HP2 <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> To figure out how many wheels needed in a large quantity of tricycle: <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> 640 tricycles <span style="font-family: 'Times New Roman','serif'; font-size: 14pt;"> 640 FSW3HP2 * 3W/ 1 FSW3HP2 <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> = 1920 W


 * [[image:TrikeEquation3.jpg width="408" height="306" caption="the bottom equation shows how to balance the top one by adding coefficients "]] ||
 * the bottom equation shows how to balance the top one by adding coefficients ||

<span style="font-family: 'Times New Roman','serif'; font-size: 12pt; margin: 0in 0in 0pt;"> Tricycle Example--Heather Bowditch


 * [[image:Drysol.jpg width="412" height="306" caption="apply the trike example to chlorine and aluminum molecules"]] ||
 * apply the trike example to chlorine and aluminum molecules ||

Using Balanced Chemical Equations **<span style="font-family: 'Times New Roman',serif; font-size: 12pt;">– Maggie Bie – pg 354 **

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> Balancing equations are used in everyday life, especially in factories making sure the cost does not exceed the profit. __<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> Key concept: __ **<span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction **

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> If the quantity of one substance is known in a reaction, you can calculate the quantity of any other substance used in the reaction <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> - Stoichiometry: The calculation of quantities in chemical reactions is a subject of chemistry. <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> - Calculations using balanced equations are called stoichiometric calculations. <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> - Used to track reactants and products using ratios of moles or representative particles.

__Practice Example (#1. bottom of page 355)__ 1. Tiny Tike has decided to make 288 tricycles each day. How many tricycle seats, wheels, and pedals are needed?
 * Heather Bowditch: **

Known: - number of trikes= 288 trikes= 288 FSW3 HP2 (see above)

Unknown: - number of seats, wheels, and pedals - seats--> S, wheels--> 3W, pedals--> 2P (for one trike)

Calculate: 288 FSW3HP2 X ( __S/ FSW3HP2)__ = 288 seats

288 FSW3HP2 X ( __3W/ FSW3HP2__ ) = 864 wheels

288 FSW3HP2 X ( __2P/ FSW3HP2__ ) = 576 pedals

ANSWERS: 288seats. 864 wheel. 576 pedals.

Interpreting Chemical Equations
 * Mike McShane (356-357) **
 * Once an equation is balanced, you can tell the relative amounts of reactants and products of the reaction that took place. A balanced chemical equation can be interpreted by either quantities such as number of atoms, molecules, or moles, and mass or volume. **

This is the chemical equation of the reaction of ammonia taking place. Refer back to this when looking at the examples: N 2 ( g ) + 3H 2 ( g ) → 2 NH 3 ( g )
 * [[image:http://upload.wikimedia.org/wikipedia/commons/9/96/Ammonia-2D-dot-cross.png width="293" height="209" caption="external image Ammonia-2D-dot-cross.png"]] ||
 * external image Ammonia-2D-dot-cross.png ||
 * <span style="color: #000000; font-family: Times New Roman; font-size: 15pt; line-height: 16px; margin: 0in 0in 10pt;"> Picture of Atom of ammonia -MikeMcShane **

When looking at the number of atoms, you can tell that the number of and types of atoms are not changed in a reaction. Ex: In ammonia, the reactants are two atoms of nitrogen, and six hydrogen atoms. In ammonia, the product has eight atoms. When looking at the number of molecules, you can tell that a ratio is constant for no matter how much of the substance you have. Ex: In ammonia, there will always be a ratio of 1:3:2 (nitrogen molecules:hydrogen molecules:molecules of ammonia). So if there are 10 molecules of nitrogen, there will always be 30 hydrogen molecules which combine to form 20 molecules of ammonia. When looking at a balanced equation, you can tell the number of moles of reactants and products. You know this by looking at the coefficients. Ex: In ammonia, the coefficient of nitrogen is 1, the coefficient of hydrogen is 3 and the coefficient of ammonia is 2. There is no ration like when looking at the number of molecules, but this tells us the amounts of reactants and products. It shows that when there is one mole of nitrogen, there is three moles of hydrogen that form two moles of ammonia. The law of conservation of mass (no mass can be created nor destroyed in a chemical or physical reaction) comes up when balancing equations. This is easy to prove because if the mass of the reactants equals the mass of the products, then you know that the equation is balanced. Ex: In ammonia, there is 1 mole of nitrogen and 3 moles of hydrogen in the reactants, and two moles of ammonia in the product. The one mole of nitrogen equals 28g (there are 2 molecules of nitrogen both at around 14g for a mass), the 3 moles of hydrogen equal 6g (there are 6 molecules of nitrogen because of the subscript 2 and the coefficient 3 all at around 1g for a mass), and the product (ammonia) has a mass of 34g. We can also look at the volume. This is the same concept as mass where the amount of volume in the reactants always equals the amount of volume in the product of a balanced equation.

Mass Conservation in Chemical Reactions In every single chemical reaction, the amount of mass in the equation and the amount of atoms in the equation are conserved. There are no equations where there will be less mass or amount of atoms in the product than the reactants and vice-versa. Molecules, formula units, moles and volumes may stay the same between the product and the reactants, but they might not. There is no guarantee that they will stay the same. 

__Practice Problem (#9 on page 358)__ Interpret the given equation in terms of relative numbers of representative particles, numbers of moles, and masses of reactants and products.
 * Heather Bowditch **

2K(s) + 2H₂O(l) à 2KOH(aq) + H₂(g)

Interpret: 2 molecules 2K + 2 molecules H₂O à 2 molecules KOH + 1 molecule H₂   2 mol 2K + 2 mol H₂O à 2 mol KOH + 1 mol H₂

**you can already tell that this equation isn’t balanced because on the left side there are 4 molecules of K, and on the right side there are only 2 molecules of K.**

Now multiply the number of moles of each reactant and product by its molar mass. 2 mol 2K + 2 mol H₂O à 2 mol KOH + 1 mol H₂

(2 mol 2K * __78.2 g/ mol__ ) + (2mol H₂O * __18.02g/mol__ ) à (2mol KOH X __56.11g/mol__ ) + (1mol H₂ X __2.02g/mol__ )

156.4 g + 36.04 g à 112.22 g + 2.02g 192.44g à 114.24g

** Conclusion: this equation is not balanced because the masses of products and reactants differ. **

12.2 Chemical Calculations
 * __ Group 5 (pgs.359-367) __**
 * By Courtney Gareau and Steven Denison **

__Courtney Gareau – Pages 359-362 –__ Writing and Using Mole Ratios __Key Concept__ – In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products.

__Mole Ratio__ – a conversion factor derived from coefficients of a balanced chemical equation interpreted in terms of moles.

Three mole ratios can be derived from the balanced equation for the production of ammonia from nitrogen and hydrogen: N2(g) + 3H2(g) à 2NH3(g) The three ratios are: Mole-Mole Calculations:

Basic Formula where W is the unknown quantity, and a & b are the coefficients from the balances equation:

Youtube Video - Courtney Gareau

Steps in Solving a Mass-Mass Problem mass G x 1 mol G / molar mass G = mol G mol G x b mol W / a mol G = mol W  mol W x molar mass W / 1 mol W = mass W
 * 1) Change the mass of G (any given mass) to moles of G (mass G à mol G) by using the molar mass of G
 * 1) Change the moles of G to moles of W (any wanted mass) (mol G à mol W) by using the mole ratio from the balanced equation.
 * 1) Change the moles of W to grams of W (mol W à mass W) by using the molar mass of W

Youtube Video - Courtney Gareau

Steve Denison - Page 363-367
Other Stoichiometric Calculations - Stoichiometry can be used to relate volumes of reactants and products in a give equation. - In a typical stoichiometric equation: - - The given quantity is first converted to moles - - Then the mole ratio from the equation (balanced) is used to calculate moles of products - - Finally the moles are convert to whatever unit of measurement the problem requires

- Mole can also be related to other quantities - - 1 mol = 6.02 x 10^23 particles - - 1 mol of gas = 22.4 L at STP (Standard Temperature and Pressure)

- Coefficents in a equation indicate relative number of particles and the relative number of moles of reactants and products - In equations with gases the coefficients also indicate relative amount of each gas - Therefore you can use volume ratios just as you would use mole ratios

Mole conversion factors: 1 mol / molar mass molar mass / 1 mol 1 mol / 6.02 x 10^23 particles 6.02 x 10^23 particles / 1 mol 1 mol / 22.4 L 22.4 L / 1 mol

SAMPLE PROBLEM 12.4
**How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation?** **2H** 2 **O(l) -> 2H** 2 **(g) + O** 2 **(g)**

1. Analyze
**Knowns** - mass of water = 29.2 g H 2 O - 2 mol of H 2 O = 1 mol O 2 - 1 mol of H 2 O = 18.0 g of H 2 O - 1 mol of O 2 = 6.02 x 10^23 molecules of O 2

**Unknown** - molecules of Oxygen = ? molecules of O 2 Therefore the following calculations must be done: g H 2 O --> mol H 2 O --> mol O 2 --> molecules O 2

2. Calculate
**Solve for the unknown** 22.92g H 2 O x (1 mol H 2 O / 18.0 g H 2 O) x (1 mol O 2 / 2 mol H2O) x (6.02 x 10^23 molecules O 2 / 1 mol O 2 ) = 4.88 x 10^23 molecules O 2

3. Evaluate
**Does this result make sense?** The given mass of water should end up with less than 1 mol of oxygen, or a little less than Avogadro's number of molecules. The answer should also have three significant figures.

SAMPLE PROBLEM 12.5
**Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog.** **How many liters of nitrogen dioxide are produced when 34 L of oxygen reacts with an excess of nitrogen monoxide? Assume conditions of STP.** **2NO(g) +** **O 2 ** **(g) ---> 2NO** 2 **(g)**

1. Analyze
**Knowns** - volume of oxygen = 34.0 L O 2 - 2 mol NO 2 / 1 mol O 2 - 2 mol O 2 ﻿= 22.4 L O 2 - 1 mol NO 2 = 22.4 L NO 2

**Unknown** - volume of nitrogen dioxide = ? L NO 2

2. Calculate
**Solve for the unknown** 34 L O 2 x (1 mol O 2 / 22.4 L O 2 ) x (2 mol NO 2 / 1 mol O 2 ) x (22.4 L NO 2 / 1 mol NO 2 ) = 68 L NO 2

3. Evaluate
**Does this result make sense?** Because 2 mol NO2 is produced for every 1 mol O2 that reacts, the volume of NO2 should be twice the given volume of O2. The answer should have two significant figures

SAMPLE PROBLEM 12.6
**Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation?** 2SO 2 (g) + O 2 (g) --> 2SO 3 (g)

1. Analyze
**Knowns** - volume of sulfur trioxide = 20.4 mL - 2 mL SO 3 / 1 mL O 2

**Unknown** - volume of oxygen = ? mL O 2

2. Calculate
**Solve for the unknown** 34 L SO 3 x (1 mol O 2 / 2 mL SO 3 ) = 10.2 mL O 2

3. Evaluate
**Does this result make sense?** Because the volume ratio is 2 volumes SO 3 to 1 volume O 2, the volume of O 2 should be half the volume of SO 3. The answer should have three significant figures.


 * __ Group 6 (pgs.368-375) __**
 * By Katherine Perry and Andrew Sciotti **

<span style="color: #ff0000; font-size: 16pt; font-style: normal; line-height: 0.2in; margin-bottom: 0in; orphans: 2; padding: 0in; widows: 2;"> Limiting and Excess Reagents
 * <span style="color: #000000; font-family: arial,helvetica,sans-serif; font-size: 12pt; font-style: normal; line-height: 0.2in; margin-bottom: 0in; orphans: 2; padding: 0in; widows: 2;"> Katherine Perry - Page 368-371 **
 * <span style="color: #000000; font-family: arial,helvetica,sans-serif; font-size: 9pt; font-style: normal; line-height: 0.2in; margin-bottom: 0in; orphans: 2; padding: 0in; widows: 2;"> Key Concept: In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. **
 * <span style="color: #000000; font-family: arial,helvetica,sans-serif; font-size: 9pt; font-style: normal; line-height: 0.2in; margin-bottom: 0in; orphans: 2; padding: 0in; widows: 2;"> __Limiting Reagent -__ reagent that determines the amount of product that can be formed by a reaction. The reaction occurs only until the limiting reagent is used up. **
 * <span style="color: #000000; font-family: arial,helvetica,sans-serif; font-size: 9pt; font-style: normal; line-height: 0.2in; margin-bottom: 0in; orphans: 2; padding: 0in; widows: 2;"> __Excess Reagent -__ reactant that is not completely used up in a reaction. **

Determining the Limiting Reagent in a Reaction
 * Copper reacts with sulfur to form copper(I) sulﬁde according to the following balanced equation. **
 * 2Cu(s) + S(s) ---> Cu2S(s) **
 * What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? **
 * 1. Analyze List the knowns and the unknowns. **
 * Knowns **
 * • mass of copper = 80.0 g Cu **
 * • mass of sulfur = 25.0 g S **
 * Unknown **
 * • limiting reagent = ? **
 * The number of moles of each reactant must ﬁrst be found: **
 * g Cu ---> mol Cu **
 * g S ---> mol S **
 * The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant: **
 * mol Cu ---> mol S **
 * The mole ratio relating mol S to mol Cu from the balanced chemical equation is 1 mol S/2 mol Cu. **
 * 2. Calculate Solve for the unknown. (pg. 370 in text) **


 * 80.0 g Cu x 1 mol Cu / 63.5 g Cu = 1.26 mol Cu **


 * 25.0 g S x ** **1 mol S / 32.1 g S = 0.779 mol S**


 * 1.26 mol Cu x ** **1 mol S / 2 mol Cu = 0.630 mol S**


 * Comparing the amount of sulfur needed (0.630 mol S) with the given amount (0.779 mol S) indicates that sulfur is in excess. Thus copper is the limiting reagent. **
 * 3. Evaluate Do the results make sense? **
 * Since the ratio of the given mol Cu to mol S was less than the ratio (2:1) from the balanced equation, copper should be the limiting reagent. **


 * Another Problem can be found on page 371 **


 * -IMAGE - Katherine Perry **


 * Percent Yield **
 * <span style="font-family: 'Times New Roman',serif; font-size: 16pt; line-height: 115%;">(Pages 372-375) Andrew Sciotti **


 * Key Concept – The percent yield is a measure of the efficiency of a reaction carried out in the laboratory**


 * Theoretical Yield** – Maximum amount of product that could be formed from given amounts of reactants
 * Ex: A cook wants to make a plate containing one piece of steak, 5 slices of potatoes, and 4 pieces of broccoli. With 10 pieces of steak, 50 pre-sliced potatoes, and 40 pieces of broccoli, the cook’s theoretical yield is 10 plates (has enough materials to make 10 plates)


 * Actual Yield** – The amount of product that actually forms when a reaction is carried out in the laboratory
 * Ex: While that same cook is trying to get 10 plates of steak, potatoes, and broccoli, a kitchen fire happens and 3 steaks are overcooked, unable to be used. Because of this, the cook can only make 7 plates, but has excess potatoes and broccoli.


 * Percent Yield** – Ratio of the actually yield to the theoretical yield, displayed as a percent
 * Percent Yield = (actually yield / theoretical yield) x 100
 * Ex: That cook that wanted 10 plates of steak, potatoes, and broccoli could only make 7 plates, so his percent yield would be: 7 (actual) / 10 (theoretical) times 100 which is 70% percent yield. That cook only produced 70% of what he could have.



<span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 18px; margin: 0px; padding: 0px;"> [| http://www.chem.arizona.edu/~salzmanr/103a004/nts004/l08/Image48.gif]

Finding theoretical yield:

What is the theoretical yield of CaO if 24.8 g CaCO3 is heated?

CaCO3 à CaO + CO2

Known: mass of calcium carbonate = 24.8 g <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> 1 mol CaCO3 = 100.1 g <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> 1 mole CaO = 58.1g <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> 24.8 g CaCO3 x (1 mol CaCO3 / 100.1 g CaCO3) x (1 mol CaO / 1 mol CaCO3) x (56.1 g CaO/ 1 mol CaO) <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> = 13.9 g aO

media type="youtube" key="TKNxdL7DN1I" height="390" width="480"

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> [| http://www.chem.arizona.edu/~salzmanr/103a004/nts004/l08/Image48.gif] <span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 115%;"> *Image and Video By Andrew Sciotti*