Chemical+Equations

=Chemical Reactions and Stoichiometry = Editor: Elena Conroy
 * A chemical reaction is a change in which one or more reactants are change into one or more products; characterized by** **the breaking of bonds in reactants and the formation of bonds in products. By the law of mass conservation, particles can not** **be made or destroyed during a chemical reaction. Stoichiometry is the portion of chemistry dealing with numerical relationships** **in chemical reactions; the calculation of quanities of substances involved in chemcial equations. In these two chapters we will****learn about chemical reactions and how they relate numerical.**

Neal McGovern (Co-Editor), Lindsey Chou, Emily Stewart
 * ﻿Describing Chemical Reactions **

**Lindsey Chou** **Pages 320 – 321** · In a chemical reaction, 1 or more substances (the reactants) change into 1 or more new substances (the products) · A chemical equation is shorthand notation to describe what is happening in a chemical equation Word Equations · Reactants and products are separated by an arrow, which means “yields” · Reactants, separated by + signs if needed, are written on the left · Products, separated by + signs if needed, are written on the right i.e. “Reactants à  products” i.e. “Methane + oxygen à  carbon dioxide + water” i.e.:
 * Writing Chemical Equations **

(Lindsey Chou.) The single man and the couple are the reactants, and they yield a different couple and a lonely man. This is an example of a single replacement reaction.


 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Neal McGovern **
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Pages 320 – 321 **
 * <span style="color: #0058ff; font-family: 'Times New Roman',Times,serif; font-size: 16pt; line-height: 150%; margin: 0in 0in 10pt;">Chemical Equations **
 * It is easier to use chemical equations using formulas than just word equations
 * A chemical equation is a representation of a chemical reaction
 * A skeleton equation is a chemical equation that is not balanced
 * In a chemical equation, you write the formulas for the reactants on the left of the yield arrow and the formulas for the products on the right
 * A catalyst is something that speeds up a chemical reaction.

this is rusty iron. The chemical formula for rust is Fe + O2 Fe2O3

the balanced chemical formula for rust is 4Fe + 3O2 2Fe2O3


 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Emily Steward **
 * Pages 324 – 328**

<span style="color: #0058ff; font-family: 'Times New Roman',Times,serif; font-size: 16pt; line-height: 150%; margin: 0in 0in 10pt;">Balancing Chemical Equations

 * To write a balanced chemical equation, first write the skeleton equation. Then use coefficients to balance the equation so that it obeys the law of conservation of mass.
 * In a balanced equation, each side has the same number of atoms of each element and mass is conserved.
 * Coefficients are small whole numbers placed in front of the formulas in an equation so it can be balanced.

1. Determine the formulas for all reactants and products. 2. Put the reactants on the left and the products on the right, separated by plus signs if necessary. Put a yields symbol (horizontal arrow) to separate the reactants from the products.
 * How to write & balance equations:**
 * 3. Figure out how many atoms there are of each element of each side. If a poly appears on both sides, treat it as a single unit.**
 * 4. Balance the elements one at a time, using coefficients. DON'T change the subscripts. It might help to start by balancing the elements that appear only once on each side of the equation.**
 * 5. Check to make sure the amount of atoms/polys is equal on both sides.**
 * 6. Can you reduce? Make sure the coefficients are in the lowest possible ratio.**


 * Example of a balanced equation:**
 * Hydrogen and oxygen react to form water.**


 * In this equation, the 2 in front of the hydrogen and the 2 in front of the water are coefficients.**
 * There are 4 hydrogen atoms and 2 atoms of oxygen on each side, so the equation is balanced.**

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

Types of Chemical Reactions Coeditor: Bryan Dextradeur

Group Members: Nikki Sheehan (Pages 330-332), Hannah Kumlin (Pages 333-335), Bryan Dextradeur (Pages 336-339).
Classifying Reactions · Five types of chemical reactions – combination, decomposition, single-replacement, double-replacement, and combustion Combination Reactions · A chemical change in which two or more substances react to form a single new substance · Magnesium and oxygen combine to form magnesium oxide 2Mg(s) + O2(g) -> 2MgO(s) · The product in this reaction is a single substance – just like in all combination reactions · More than one product is possible when two nonmetals combine S­­­(s) + O2(g) -> SO2(g) sulfur dioxide 2S(s) + 3O2(g) -> 2SO3(g) sulfur trioxide · Also more than one product is possible when a transition metal and a nonmetal combine Fe(s) + S(s) -> FeS(s) iron(II) sulfide 2Fe(s) + 3S(s)­ -> Fe2S3(s) iron(III) sulfide Decomposition Reactions · A chemical change in which a single compound breads down into two or more simpler products · There is only one reactant but two or more products · Most decomposition reactions require energy in the form of heat, light, or electricity 2HgO(s) -> 2Hg(l)­ + O2(g)
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Nikki Sheehan **
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Pages 330– 332 **

Nikki Sheehan

**Single-Replacement Reactions**

 * Dropping a small piece of potassium into a beaker of water creates a vigorous reaction, as shown in the picture below:

Hannah Kumlin


 * This reaction produces hydrogen gas and a large quantity of heat. The released hydrogen gas can ignite explosively. You would write this reaction as

2K + 2H2 O -->2
 * This equation is an example of a single-replacement reaction. A single replacement reaction is a chemical change in which one element replaces a second element in a compound. You can identify this form of reaction by noting that both the reactants and the products consist of an element and a compound. For instance in the equation above K and H are the elements and H2 O and KOH are the compounds.
 * Whether one metal will displace another metal from a compound depends upon the relative reactivities of the two metals. The activity series of metals lists metals in order of decreasing reactivity. *You can find the chart showing the activity series of metals on page 333 in your textbook*. A reactive metal will replace any metal listed below it in the activity series.
 * A halogen can also replace another halogen from a compound. The activity of the halogens decreases as you go down Group 7A of the periodic table.


 * for single-replacement reaction problems see the top of page 334.**

Double-Replacement Reactions
Hannah Kumlin
 * notice their hats*


 * Sometimes, when two solutions of ionic compounds are mixed, nothing happens. At other times, the ions in the two solutions react. A ** double-replacement reaction ** is a chemical change involving an exchange of positive ions between two compounds. These reactions are also referred to as double-displacement reactions. They generally take place in an aqueous solution and often produce a precipitate, a gas, or a molecular compound such as water. For these reactions to occur, one of the following is generally true:
 * 1) One of the products is only slightly soluble and precipitates from solution.
 * 2) One of the products is a gas.
 * 3) One product is a molecular compound such as water.

media type="youtube" key="MkXm7xu8HyI" height="390" width="480" Hannah Kumlin


 * for double-replacement practice problems, see page 335.**

Pages 336-339

 * Combustion Reactions **
 * 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.
 * Examples are campfires and gas grills.
 * Hydrocarbon, a compound consisting of hydrogen and carbon, is often the other reactant in Combustion Reactions.
 * The complete combustion of a hydrocarbon produces carbon dioxide and water.
 * If the supply of oxygen is limited during the reaction, the combustion will not be complete.
 * Elemental Carbon (Soot) and toxic Carbon Monoxide gas may be additional products of combustion.
 * The complete combustion of a hydrocarbon releases a large amount of energy as heat.
 * This is why hydrocarbons such as Methane (CH4﻿), Propane (C3H8﻿), and Butane (C4H10﻿) are important fuels.
 * Gasoline is a mixture of hydrocarbons that can be represented approximately by the equation (C8H18﻿).
 * The reactions between oxygen and some elements other than Carbon are also examples of combustion reactions.
 * For example, both magnesium and sulfur burn in the presence of oxygen.
 * The following equations are examples of combustion reactions. Notice that they can also be classified as combination reactions.
 * 2Mg+O2﻿ -> 2MgO
 * S+O2﻿ -> SO2

= Predicting the Products of a Chemical Reaction = media type="youtube" key="lPvqk5OwtDs" height="390" width="480"
 * The number of elements and/or compounds reacting is a good indicator of possible reaction type and thus possible products.
 * 1. Combination Reaction
 * General Equation: R+S -> RS
 * Reactants: Generally two elements, or two compounds (where at least one compound is a molecular compound).
 * Probable Products: A Single Compound.
 * Example: Burning Magnesium in air.
 * 2. Decomposition Reaction
 * General Equation: RS -> R+S
 * Reactants: Generally a single binary compound or a compound with a polyatomic ion.
 * Probable Products: Two elements (for a binary compound), or two or more elements and/or compounds (for a compound with a polyatomic ion).
 * Example: Heating Mercury (II) Oxide.
 * 3. Single-Replacement Reaction
 * General Equation: T+RS -> TS+R
 * Reactants: An element and a compound. In a single-replacement reaction, an element replaces another element from a compound in aqueous solution. For a single-replacement reaction to occur, the element that is displaced must be less active than the element that is doing the displacing.
 * Probable Products: A different element and a new compound.
 * Example: Potassium in water.
 * 4. Double Replacement Reaction
 * General Equation: R+S- + T+U- ﻿-> R + U- + T+S-
 * Reactants: Two ionic compounds. In a double-replacement reaction, two ionic compounds react by exchanging cations to form two different compounds.
 * Probable Products: Two new compounds. Double-Replacement reactions are driven by the formation of a precipitate, a gaseous product, or water.
 * Example: Reaction of aqueous solutions of barium chloride and potassium carbonate.
 * 5. Combustion Reactions
 * Equation: Cx﻿Hy ﻿+ (x+y/4)O2﻿ -> xCO2 + (y/2)H2﻿O.
 * Reactants: Oxygen and a compound of C, H, (O). When oxygen reacts with an element or compound, combustion may occur.
 * Probable Products: CO2 and H2O. With incomplete combustion, C and CO may also be products.
 * Example: The combustion of methane gas in air.

Above Image: Bryan Dextradeur

<span style="color: #008000; font-family: 'Times New Roman',Times,serif; font-size: 140%;">Reactions In Aqueous Solution
__**<span style="font-family: 'Times New Roman',Times,serif; font-size: 120%;">By: David Monti, Lauren Murphy, and Chris Kelly (Co-Editor) **__


 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">David Monti **
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Page 342 **

<span style="color: #008000; font-family: 'Times New Roman',Times,serif; font-size: 130%;">Net Ionic Equations
<span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Ag + NO3 + Na + Cl -> AgCl + Na + NO3
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Many important chemical reactions take place in water, aqueous solution
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Ex. The aqueous solutions of silver nitrate with sodium chloride to for silver chloride and aqueous sodium nitrate (double replacement reaction)
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">This standard method of writing equations does not show that like most ionic compounds, the reactants and one of the products separate into cations and anions when they dissolve in water
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Complete ionic Equation – an equation that shows dissolved ionic compounds as dissociated free ions
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Ions that remain unchanged throughout the entire chemical reaction, and are the same in the product as they are in the reactants are not required in the equation

<span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">**<span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Lauren Murphy ** <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">﻿<span style="color: #008000; font-family: 'Times New Roman',Times,serif; font-size: 130%;">Net Ionic Equations (cont.) <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Ag + Cl > AgCl
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Page 343 **
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Spectator ion – an ion that appears on both sides on an equation and is not directly involved in the reaction
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">Net ionic equation – an equation for a reaction in solution that shows only those particles that are directly involved in the chemical change
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">When removing the unnecessary ions to make a net ionic equation, one must make sure that the charge of the equation remains balanced
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">A net ionic equation shows only those particles involved in the reaction and is balanced with respect to both mass and charge

<span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;"><span style="font-family: 'Times New Roman',serif;"> **Chris Kelly** <span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">**<span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 24px; margin-bottom: 10pt; margin-left: 0in; margin-right: 0in; margin-top: 0in;">Page 344 ** <span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;"><span style="font-family: 'Times New Roman',serif; font-size: 12pt; line-height: 24px; margin-bottom: 10pt; margin-left: 0in; margin-right: 0in; margin-top: 0in;">﻿<span style="color: #008000; font-family: 'Times New Roman',Times,serif; font-size: 21px;">Predicting the Formation of a Precipitate
 * mixing solutions of two ionic compounds can result in formation of insoluble salt (Precipitate)
 * Sometimes this is not the case
 * It all depends on solubility
 * Formation of precipitate can be predicted by general rules
 * Salts of alkali metals and ammonia = soluble
 * Nitrate salts and chlorate salts = soluble
 * Sulfate salts, except compounds with Pb 2+, Ag +, Hg2 2+, Ba 2+, Sr 2+, and Ca 2+ = Soluble
 * Chloride salts, except compound with Ag +, Pb 2+, and Hg2 2+ = Soluble
 * Carbonates, phosphates, chromates, sulfides, and hydroxides = Most are insoluble

**<span style="font-family: 'Times New Roman',Times,serif;">The Arithmetic of Equations **
 * Allie Chabot: Co Editor (Pg. 353-355)**
 * Ellie Kawa (356-358)**


 * Allie Chabot**
 * 12.1 The Arithmetic of Equations**


 * Using Everyday Equations **
 * A balanced chemical equation provides the same kind of quantitative information that a recipe does.
 * A balanced equation includes the reactants or the “recipes”, and the products or the “formula”.

(Allie Chabot)
 * Using Balanced Chemical Equations **
 * Nearly everything we use in our daily life is manufactured from chemicals.
 * Balanced chemical equations tell us the amount of reactants to mix and what amounts of products to expect.
 * Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.
 * When one knows the quantity of one substance in a reaction, one can calculate the quantity of any other substance consumed or created in the reaction.
 * Quantity usually means the amount pf a substance expressed in either grams, moles, liters, tons, or molecules.
 * Stoichiometry- that portion of chemistry dealing with numerical relationships in chemical reactions; the calculation of quantities of substances involved in chemical equations.
 * Ellie Kawa**
 * Pgs 356-358**


 * Interpreting Chemical Equations **
 * · a balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume
 * at the atomic level, a balanced equation indicates that the number and type of each atom that makes up each reactant also makes up each product
 * o both the number and types of atoms are not changed in a reaction
 * nitrogen and hydrogen will always react to form ammonia in a 1:3:2 ratio of molecules
 * o N2(g)+3H2(g)®2NH3(g)
 * § if you could make 10 molecules of nitrogen react with 30 molecules of hydrogen, you would expect to get 20 molecules of ammonia
 * § you could take Avogadro’s number of nitrogen molecules and make them react with three times Avogadro’s number of hydrogen molecules- there would still be the same 1:3 ratio of molecules of reactants
 * the reaction would form two times Avogadro’s number of ammonia molecules
 * a balanced chemical equation tells you the number of moles of reactants and products
 * o the coefficients of a balanced chemical equation indicate the relative numbers of moles of reactants and products in a chemical reaction
 * § this is the most important information that a balanced chemical equation provides[[image:http://balancingequations.info/Images/water-equation-add-water-on-right.jpg align="right" caption="Ellie Kawa"]]
 * § using this information you can calculate the reactants and products
 * a balanced chemical equation obeys the law of conservation of mass
 * o this law states that mass can neither be created nor destroyed in an ordinary chemical or physical process
 * o using the mole relationship, you can relate mass to the number of atoms in the chemical equation
 * if you assume standard temperature and pressure, the equation tells you about the volume of gases

Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is: 2H2S(g)+3O2(g)®2SO2(g)+2H2O(g)
 * Mass Conservation in Chemical Reactions **
 * · mass and atoms are conserved in every chemical reaction
 * o however, molecules, formula units, moles, and volumes are not necessarily conserved- although they may be
 * Conceptual Problem **
 * Interpreting a Balanced Chemical Equation **

Interpret this equation in terms of a. numbers of representative particles and moles b. masses of reactants and products Analyze a. the coefficients in the balanced equation give the relative number of molecules or moles of reactants and products b. a balanced chemical equation obeys the law of conservation of mass

Solve

a. 2 moleculesH2S+ 3molecules O2®2 moleculesSO2+ 2 moleculesH2O 2 mol H2S+ 3 mol O2®2 mol SO2+ 2 mol H2O

b. multiply the number of moles of each reactant and product by its molar mass: 2 mol H2S+ 3 mol O2®2 mol SO2+ 2 mol H2O (2 mol*34.1g/mol)+(3 mol*32.0g/mol)®(2mol*64.1g/mol)+(2 mol*18.0g/mol) 68.2 g H2S+ 96.0 g O2®128.2 g SO2+ 36.0 g H2O 164.2 g=164.2 g


 * Chemical Calculations **
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Group 5: Shannon Leavey(coediter) and Jessen Foster **

<span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">**Shannon Leavy**
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Pages 359– 362 **

Writing and Using Mole ratios:
-mole ratio: a conversion factor derived from coefficients of a balanced chemical equation interpreted in terms of moles -In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of rectants, or between moles of products example: __1mol N2__ 3 mol H2


 * Steps in Solving a Mass-Mass Problem: **

1.Change the mass of G to moles of G (mass G> mol G) by using the molar mass of G. mass G x __1 mol G__ = mole G molar mass G

2. Change the moles of G to moles of W (mol G---> mol W) by using the mole ratio from the balanced equation. mol G x __//b// mol W__= mol W //a// mol G 3. Change the moles of W to grams of W (mol W---> mass W) by using the molar mass of W. mol W x __molar mass W__ = mass W 1 mol W media type="youtube" key="2Q8HtHItjxI" height="390" width="480" <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">**Jessen Foster**
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Pages 363– 367 **

** Other Stoichiometric Calculations **
-in a typical stoichiometric problem: how many molecules of oxygen are produced when 29.2 grams of water is disposed by electrolysis according to this equation: 2H2O → 2H2 + O2 29.2 gH2O X 1 mol H2O X 1 mol O2 X 6.02x10^23 molecules of O2 18.0 gH2O 2 mol H2O 1 mol O2 = 4.88x10^23 molecules O2
 * 1) the given quantity is first converted to moles
 * 2) the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance
 * 3) moles are converted to any other unit of measure relating to the unit
 * -calculate molecules of a product**

nitrogen monoxide and oxygen gas combine to form brown gas nitrogen dioxide, which contributes to photochemical smog: how many liters of nitrogen dioxide are produced when 34 liters of oxygen reacts with an excess of of nitrogen monoxide?: 2NO + O2 → 2NO2 34 L O2 X 1 mol X 2 mol NO2 X 22.4 L NO2 22.4 L O2 1 mol O2 1 mol NO2 = 68 NO2 -finding the volume of a gas needed for a reaction assuming STP, how many milliliters of oxygen are needed to produce 20.4mL SO3 according to this equation: 2SO2 + O2 → 2SO3 20.4mL SO3 X 1 mL O2 2mL SO3 = 10.2mL O2
 * -volume-volume stoichiometric calculations**


 * Limiting Reagent and Percent Yield **
 * <span style="font-family: 'Times New Roman','serif'; font-size: 12pt; line-height: 150%; margin: 0in 0in 10pt;">Group 6: ****<span style="font-family: 'Times New Roman',Times,serif; font-size: 12pt;">coeditor- kayla anghinetti & teresa lynch **


 * Teresa Lynch **
 * Pages 368– 371 **

Limiting and Excess Reagents:
> Solving: knowns: mass of Cu= 80.0g Cu and mass of S= 25.0g S > unknown: limiting reagent= ? > 1. find number of moles (gCu--> moles Cu and gS--> moles S) > 2. balanced equation used to find # moles of 1 reactant to another (molCu --> molS) > 3. mole ratio relates mol S to mol Cu from balanced equation is 1 molS/ 2mol Cu > 4. solve: 80.0gCu X 1molCu/ 63.5 gCu= 1.26 molCu > 25.0gS X 1molS/ 32.1 gS= 0.779 molS > 1.26molCu X 1mol S/2mol Cu= 0.630 molS > > media type="youtube" key="dAkbcRgecBo" height="390" width="480" Picture of copper sulfide, the product of the sample equation above. (T.Lynch)
 * In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.
 * A chemical equation can be interpreted on a microscopic (particles) or macroscopic ( moles) scale.
 * Limiting reagent- reagent that determines the amount of product that can be formed by a reaction.
 * The reaction can occur until the limiting reagent is used up.
 * Excess reagent- is the reactant that is not completely used up in a reaction.
 * If the chemical equation in stoichiometry is not in moles it must first be converted into moles, next you can identify the limiting reagent and from that determine the amount of the product there will be.
 * Example problem for finding the limiting reagent in a reaction:
 * Problem: 2Cu (s) + S(s) -->
 * Answer: comparing 0.630 molS with 0.779 molS shows that sulfur is the excess; therefore copper is limiting reagent.

You can also solve using the limiting reagent from this problem and the molar mass of 1mol Cu2S to find the mass of copper sulfide (or the quantity of the product).


 * Kayla Anghinetti **
 * Pages 372– 375 **

Percent, Actual, Theoretical Yield
I. Percent Yield a. When an equation is used to calculate the amount of product that forms during a reaction, the calculated value is the theoretical yield. -theoretical yield: maximum amount of product that could be formed from given amounts of reactants. b. Actual yield: amount of product that actually forms. c. Percent yield: ratio of the actual to theoretical yield expressed as a percent. d. Percent yield is a measure of the efficiency of a reaction carried out in the lab. SAMPLE PROBLEMS -Calculating the Theoretical Yield of a Reaction Using the balanced equation CaCO­­­­­­­­­3 à CaO + CO2, find the theoretical yield of CaO if 24.8g of CaCO3 is heated.

1. List givens - given mass of CaCO3 = 24.8g - molar mass of CaCO3 = 100.1g - molar mass of CaO = 56.1g 2. Solve for unknown (g of CaO)

-Calculating the Percent Yield of a Reaction What is the percent yield if 13.1g CaO is actually produced when 24.8g CaCO3 is heated? CaCO­­­­­­­­­3 à CaO + CO2

1. List givens -actual yield = 13.1g CaO -theoretical yield =13.9g CaO (from previous problem) -percent yield = actual/theoretical x 100%

2. Solve for unknown (percent yield)