Chemical+Reactions+&+Stoichiometry


 * **__Chapter 11 & 12: Chemical Reactions & Stoichiometry __**

=Introduction =

﻿This wiki has been constructed in order to help understand the broad subjects of chemical reactions (chapter 11) and stoichiometry (chapter 12). Chapter eleven instructs on how to write chemical equations, describing them, and identifying t he five different types of chemical reactions. It also focuses on balancing chemical equations, as well as reactions that happen in aqueous solutions. The last sections of the chapter informs readers about the formations of precipitates.

Chapter tweleve explains how balanced equations apply to both chemistry and everyday life. In addition, balanced chemical equations are interpreted in terms of moles, representative particles, mass, and gas volume at STP. The calculations that are necessary for the chemical reactions are explored. Chapter tweleve also teaches how balanced chemical equations can be utilized to calculate reactants and products of a reaction. The quantities that are always conserved in chemical reactions are taught to be identified. Common vocabulary terms in relation to labs that are relative to this section are theoritical yield, percent yield, excess reagent, and limiting reagent. =Groups = __Editor__ - Gabe Hannawi

1. Describing Chemical Reactions (pp. 321-329) Co-Editor: Zac Boulerice Group Members: Alex Trombetta, Tess Murphy

2. Types of Chemical Reactions (pp. 330-339)

Group Members: Joe Hatch, Kevin McAllister
3. Reactions in Aqueous Solution (pp. 342-345) Co-Editor: Erin Cropanese Group Members: Nick Romero, Sahana Nazeer

4. The Arithmetic of Equations (pp. 353-358) Co-Editor: Cassie Namie Group Members: Henry Dodge, Andrew Ware

<span style="color: #ff0000; font-family: Arial,Helvetica,sans-serif;">5. Chemical Calculations <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">(pp. 359-367) Co-Editor: Jess Mahoney Group Members: Caitlyn Vogt

<span style="color: #ff0000; font-family: Arial,Helvetica,sans-serif;">6.Limiting Reagents & Percent Yield (pp. 368-375) Co-Editor: Ian Travis Group Members: Lauren Sacks

Content

<span style="color: #ff0000; font-family: Arial,Helvetica,sans-serif;">1. Describing Chemical Reactions <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">(pp. 321-329) Co-Editor: Zac Boulerice Group Members: Alex Trombetta, Tess Murphy

<span style="color: #0000ff; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; font-size: 12px; line-height: normal;">Zac Boulerice Pages 321-323


 * __<span style="color: #0000ff; font-family: 'Palatino Linotype','Book Antiqua',Palatino,serif; font-size: 12px; line-height: normal;">11.1: Describing Chemical Reactions __**

Writting Chemical Equations:

Key Concepts:

- How to write word equations - How to write a skeleton equation

Vocab:

- Chemical Equation: A representation of a chemical reaction - Skeleton Equation: A chemical equatation that does not represent the relative amounts of the reactants and products - Catalyst: Something that speeds up a reaction but isn't involved in it.

Word problems:

"Iron reacts with oxygen to produce iron(III) oxide (Rust)

or

Iron + Oxygen ---> Iron(III) oxide

(Just leave the reactants and products and add plus signs and an arrow where needed)

Example of a Chemical and Skeleton Equation:

Fe + O2 -> Fe2O3

(Notice the amount of Iron and Oxygen is not equal on both sides, this will be explained in the next section)

Symbols used in chemical equations

<-- || Used in place of "-->" for reversible reactions || ---> || Indicates that heat is supplied to the reaction || --> || A formula written above or below the yield sign indicates its use as a catalyst (for this example, platinum ||
 * Symbol || Explanation ||
 * + || Used to separate two reactants or products ||
 * --> || "Yields," separate reactants from products ||
 * (s) || Solid ||
 * (l) || Liquid ||
 * (g) || Gas ||
 * (aq) || Aqueous solution (dissolved in water ||
 * (Triangle)
 * Pt

Alex Trombetta Pages 324-326 = Balancing Chemical Equations =

Key Concept: To write a balanced equation first write the skeleton equation. Then use coefficients to balance the equation so that it obeys the law of conservation of mass. A chemical equation is much like a recipe. In an equation there is a list of ingredients. A word equation could look like this

Reactants Product

Eggs + Milk + Flour + Sugar = Cake

This equation however, does not tell you the quantity of each item, and without the quantity you are unable to make a cake. A general cake contains 2 eggs (E) 1 cup of milk(M) 1 cup of flour(F) and 2 cups of sugar(S). Therefore, the equation of the cake would be,

E + M + F + S= E2MFS2

This equation is unbalanced because there is not the same amount of ingredient on each side, and in order to balance it there has to be another E and another S on the reactant side.

E2 + M + F + S2= E2MFS2

For an equation to be balanced, the reactants must equal the product. In order to change the amount of the reactant, you have to add what is called a coefficient in front of what you would like to change the amount of like what was done in bold,

Coefficient: A small whole number that is placed in front of a formula in an equation in order to balance it. Balanced Equation: A chemical equation in which mass is conserved; each side of the equation has the same number of atoms of each element.

Equations must be balanced because according to Daltons theory, no new bonds are formed are destroyed but instead they are arranged in a new manor. The best strategy for balancing equation is with the use of guess and check. For example in the equation

S8 + O2= SO2

There are three spots for coefficients demonstrated by the. This equations is not balanced because the number of Sulfur on the reactant side does not equal that on the product side. To fix this you must at an 8 to the product side therefore making it

S8 + O2= 8SO2 But now the amount of oxygen is unequal on either side because there is 16 oxygen on the product side and only 2 on the reactant side. Therefore if you add and 8 to in front of the oxygen there is 16 on both sides. S8 + 8O2= 8SO2 The equation is then balanced. To check your work count the amount of each particle on each side. If there is the same amount on each side then the equation is balanced. In this equation for example there are 8 sulfur atoms on the left and right side of the equation, and there are 16 Oxygen atoms on the right and left of the equation. You also have to try to simplify if there is a common factor in your coefficients. In this case there is no common factor and the equation is balanced.

More Practice Problems Balance these equations

1.) H2 + O2 = H2O 2.) Zn + HCl = ZnCl2 + H2  3.) Pb(NO3)2 + AlCl3 = PbCl2 + Al(NO3)3

*Helpful Technique* http://www.youtube.com/watch?v=VUnYowO5_78

Tess Murphy (pg 327- 329) Watch this video below about balancing equations! Listen to a little Speak Now by T-Swift in the background. I couldn't put it in for copyright reasons. media type="youtube" key="_KDeI0zZcq0" height="390" width="480"

Try This For a Fun Way to Balance Equations! Practice Balancing Equations Learn about Hazardous Material Specialists Here!


 * <span style="display: inline !important;">Co-editor Chris Hughes pgs 333-336 **


 * <span style="display: inline !important;">Kevin McAllister 330-332 **


 * <span style="display: inline !important;">Joe Hatch 337-339 **

- Kevin McAllister (330-332) Classifying Reactions


 * Five General types of reactions are combination decomposition single replacement, double replacement and combustion
 * Occasionally reactions fall into multiple categories
 * Patterns of chemical behavior become apparent
 * Combination Reactions
 * Chemical change where two or more substances for a single new substance
 * The products will be a single compound (like MgO, not MgO + something else)
 * Patterns
 * Group A metal and nonmetal—product is a metal cation and the nonmetal anion
 * 2 nonmetals react---more than one product is possible
 * More than one product results from the combination reaction of a transition metal and a nonmetal
 * Practice Problems
 * Be + O2 à
 * Write and balance the equation for the formation of Magnesium Nitride(Mg3N2)
 * Check out this [|link] for a little more help
 * Decomposition Reactions
 * A chemical change which a single compound breaks down into two or more simpler products
 * Only one reactant and two or more products
 * HgO breaks down into Mercury and O2
 * Practice Problems
 * Complete and balance this decomposition reaction
 * HI
 * Write the formula for the binary compound that decomposes to the products H2 and Br2
 * Check out this[| video] for a little more help!
 * Good Luck
 * J

Chris Hughes (333-336) o Chemical change in which one element replaces a second element in a compound. o Example: 2K + 2H2O à 2KOH + H2 o After the reaction the potassium replaced the hydrogen o The dtetermining factor of whether one metal replaces another is based on the reactivity of the two metals. o The activity series of metals lists the metals in order of reactivity. o The reactive metal will replace any metal listed below it. o For an easy-to-follow activity series of metals, click [|here] o Also referred to as double displacement reactions o Occurs when positive ions are exchanged between two compounds o Example: K2CO3 + BaCl2 à 2KCL + BaCO3 o The potassium and the barium are the positive ions. o Go to this [|link] for an animation describing Double-Replacement Reactions o Chemical change where an element or compound reacts with oxygen o Fire is a result of a combustion reaction. o Example: CH4 + 2O2 à CO2 + 2H2O o Combustion reactions commonly involve hydrocarbons, which are compounds that consist of hydrogen and carbon.
 * Single-Replacement Reactions
 * Double-Replacement Reactions
 * Combustion Reactions

Joe Hatch (337-339) Writing Equations for Combustion Reactions Example: Write balanced equations for the complete combustion of these compounds: benzene (C6H6) and ethanol (CH3CH2OH) Step 1: products are CO2 and H2O. Write a skeleton equation for each reaction, and then balance them. Step 2: benzene: C6H6 + O2 → CO2 + H20 Balanced (benzene): 2C6H6 + 15O2 → 12CO2 + 6H2O Ethanol: CH3CH2OH + O2 → CO2 + H2O Balanced (ethanol): CH3CH2OH + 3O2 → 2CO2 + 3H2O

Predicting the Products of a Chemical Reaction -combination reaction: two or more reactants form one product R + S → RS -decomposition reaction: one reactant yields more than one product RS → R + S -Single-replacement reaction: an element and compound are reactants and a different element and compound are the reactants T + RS → TS + R -Double-replacement reaction: two ionic compounds are reactants and the products are two new compounds R(+)S(-) + T(+)U(-) → R + U(-) + T(+)S(-) -combustion reaction: oxygen and usually a hydrocarbon are the products and usually the products are carbon dioxide and water C(x)H(y) + (x + y/4)O2 → xCO2 + (y/2)H2O

** __Sahana Nazeer (344)__ **
Reactions in Aqueous Solutions (Erin Cropanese pgs. 342-343)

Net Ionic Equations -many chemical reactions take place in water(aqueous solution)

example: AgNo3(aq) + NaCl(aq)-> AgCl(s) = NaNO3(aq) complete ionic equation: an equation that shows dissolved ionic compounds as dissociated free ions
 * this equation does not show that like most ionic compounds the reactants, and one of the products dissociate, or separate, into cations and anions when they dissolve in water
 * when sodium chloride dissolves in water it separates into sodium
 * silver nitrate dissociates into silver ions and nitrate ions
 * you can use these ions to write a complete ionic equation

Ag(aq)+ No3-(aq) + Na(aq) + Cl- (aq) > AgCl(s) + Na+(aq) + No3-(aq) *notice the nitrate ion and the sodium ion appear unchanged on both sides of the equation *eliminate the ions that don’t participate in the reaction

Ag(aq)+ No3-(aq) + Na(aq) + Cl- (aq) > AgCl(s) + Na+(aq) + No3-(aq)

spectator ion:** an ion that appears on both sides of an equation and is not directly involved in the reaction


 * leave out the spectator ions in an equation and you will have the net ionic equation

net ionic equation: an equation for a reaction in solution that shows only those particles that are directly involved in the chemical change

example: the reaction of lead with silver nitrate
 * when writing a balanced net ionic equation, make sure that the ionic charge is balanced

Pb(s)+ AgNO3 (aq)>Ag(s)+ Pb(NO3)2 (aq)

*The nitrate ion is the spectator ion in this reaction. The net ionic equation is this:

Pb(s)+ Ag+(aq)>Ag(s)+ Pb+2 (aq) [unbalanced]

*notice that a single unit of positive charge is on the reactant side of the equation. Two units of positive charge are on the product side. Placing the coefficient 2 in front of Ag+(aq) balances the charge. A coefficient of 2 in front of Ag(s) rebalances the atoms.

Pb(s)+ 2Ag+(aq)>2Ag(s)+ Pb+2 (aq)

media type="youtube" key="MeSi3dDOL2I" height="390" width="480"

Predicting the Formation of a Percipitate (Sahana Nazeer 344)

Whether or not a percipitate forms depends on the solubility of the new compounds that form. Hence, you can predict the formation of a percipitate by using the general rules for solubility of ionic compounds.

Nitrates salts and chlorate salts -- soluble Most sulfate salts -- soluble Most chloride salts -- soluble Carbonates, phosphates, chromates,sulfides, and hydroxides -- most insoluble
Solubility Chart: @http://www.professormeyer.com/images/solubility%20chart.jpg

When you mix two solutions of different ionic compounds that are individually soluble compounds, then they will form an insoluble compound..which produces a precipitate.

Example. Will a precipitate form when aqueous solutions of sodium carbonate and barium nitrate are mixed?

-when the four ions mix, the cations change partners; the two new compounds resulting of this potential exchange is sodium nitrate and barium carbonate. -you must refer to whether or not the ion combinations are soluble, which indicates if an exhange will be made. -sodium nitrate will not form a percipitate because sodium is an alkali metal (it is soluble) -barium carbonate will percipitate

Below is a great website goes step by step on predicting the formation of a percipitate with practice problems and answers. @http://dl.clackamas.edu/ch105-04/predicti.htm

Predicting the Formation of a Percipitate Nick Romero- Page 345 Section Review Pretty much the everything that Sahana had covers the topic here are other helpful links I found media type="youtube" key="eDT_sOTJ2-0" height="390" width="640"

GREAT WEBSITE ALSO === http://dl.clackamas.edu/ch105-04/predicti.htm

__ Chapter 12: Stoichiometry __

The Arithmetic of Equations (Henry Dodge 353-354)

Using Everyday Equations -A balanced chemical equation gives the same kind of quantitative information as a recipe. Using Balanced Chemical Equations -Chemists use balanced chemical equations to calculate the amount of reactant needed or product that will be produced in a reaction. -If the quantity of one substance in a reaction is known, the quantities of any other substance can be calculated. Quantity is usually expressed in grams or moles. Stoichiometry – the calculation of quantities in chemical reactions - Stoichiometric calculations are calculations using balanced equations. - Stoichiometry is used to tally the amounts of reactants and products using ratios of moles or representative particles.



Cassie Naimie (definition of stoichiometry and images)

-Mass and atoms are conserved in every chemical reaction.
= 12.3:Limiting Reagents and Percent Yield = By Lauren Sachs, Ian Travis (Co editor)

Limiting and Excess Reagents (Pgs. 368-371)
By Ian Travis
 * A chemical Reaction with a certain set of reactants can continue only as long as the required reactants are present
 * For example, in the equation N2 + 3H2 ---> 2NH3 if two molecules of (N2) were to react with 3 molecules of hydrogen (2N2 + 3H2), the product of (2NH3) would remain the same, but one molecule of (NH2) would remain, all the hydrogen having been used up.
 * Because the amount of hydrogen is what determined when the reaction had to stop (the reaction requires hydrogen to take place), it becomes the limiting reagent of this equation. The left over nitrogen is the excess reagent
 * Limiting Reagent: The reagent that determines the amount of product that can be formed by a reaction
 * Excess Reagent: The reactant that is not completely used up by the reaction.
 * The concepts of Excess and Limiting Reagents are actually extremely simple. It can be illustrated in many everyday situations, such as cars and tires:
 * [[image:http://www.chem.tamu.edu/class/majors/tutorialnotefiles/cars.gif]]"Cars" would be the limiting reagent, with "Tires" as excess.
 * In Chemistry to determine the amount of excess in a reaction, samples of substances in other units such as grams must first be converted into moles, and have their equations determined.

Percent Yield (pgs. 372-375)
By Lauren Sachs
 * Chemists use calculations when the product from a chemical reaction is less than expected
 * Theoretical Yield is the maximum amount of a product that can be formed with given amounts of reactants
 * Actual Yield is the amount of product that actually results
 * Percent Yield= (actual yield/theoretical yield) x 100%
 * Actual yield is often less than theoretical yield, meaning the result is usually less than 100%
 * The percent yield is a measure of efficiency of a reaction carried out in the laboratory
 * There are many reasons percent yield is less than 100%
 * Reactions may not go all the way to completion
 * Impurity of reactants and side reactions can cause extra products
 * Some of product can be lost in filtering or transferring
 * Finally, careless measurements can affect percent yield

> > > > Actual yield is an experimental value >>>>> media type="youtube" key="CMi9k8ts9r4" height="390" width="640" align="left" > > > > > > > > > > > > > > > > > > > > > > > Sample Problem: Calculating the Theoretical Yield of a Reaction > CaCO3﻿(s) --> CaO(s) + CO2(g) : What is the theoretical yield of CaO if 24.8g CaCO3 is heated > > 1. Analyze: List knowns and unknowns > > > 2. Calculate: Solve for the unknown > 3. Evaluate: Does the result make sense? > > > > Sample Problem: Calculating the Percent Yield of a Reaction > CaCO3﻿(s) --> CaO(s)+CaCO2(g) : What is percent yield if only 13.1 g CaO is produced when 24.8g CaCO3 is heated > > 1. Analyze: List the knowns and unknowns > 2. Calculate: Solve for the unknown > 3. Evaluate: Does the answer make sense > Practice Problems > When 84.8g of iron(III) oxide reacts with an excess of carbon monoxide, iron is produced (Fe2O3(s) + 3CO(g) --> 2Fe(s) +3CO2(g)); Find the theoretical yield of reaction (the iron). > > When 5.00g of copper reacts with silver nitrate, silver and copper(II) nitrate are produced. What is the theoretical yield of silver? > > If 50.0g of silicon dioxide is heated with an excess of carbon, 27.9g silicon carbide is produced. What is the percent yield? > Hint: You must first find theoretical yield using SiO2(s) + 3C(s) --> SiC(s) + 2CO(g) > > If 15.0g of nitrogen reacts with 15.0g of hydrogen, 10.5 g of ammonia is produced. What is the percent yield? > > > [|Practice Problems]
 * Known: mass of calcium carbonate=24.8g CaCO3, 1 mole CaCO3=100.1g CaCO3, 1 mol CaO=56.1g CaO
 * Unknown: theoretical yield of Calcium Oxide= ?g CaO
 * g CaCO3﻿-->mol CaCO3﻿-->mol CaO-->g CaO; appropriate mole ratio is 1mol CaO/1mol CaCO3
 * This is because there is only 1 mole of each on each side
 * 24.8g CaCO3﻿ x (1mol CaCO3﻿/100.1g CaCO3﻿) x (1mol CaO/1mol CaCO3﻿) x (56.1g CaO/1mol CaO)= 13.9g CaO
 * The mole ratio is 1:1, and the ratio of their molar masses is slightly over 1:2
 * Calculations show that mass of CaO is slightly more than half the mass of CaCO3
 * Known: actual yield=13.1g CaO; theoretical yield=13.9g CaO (see above); and percent yield=(actual yield/theoretical yield) x 100%
 * Unknown: percent yield=?%
 * percent yield=(13.1g CaO/13.9g CaO) x 100% = 94.2%
 * The actual yield is slightly less than the theoretical yield, and 94.2% is slightly less than 100%