Group 1
David O'Brien p. 323-324 Chemical Equations Chemical Equations = a representation of a chemical reaction

The reactant is on the right, an arrow in the middle pointing towards the products on the left

Ex.) Fe + O2 à Fe2O3 Ex.) AgNO3 + H2S à Ag2S + HNO3 Skeleton Equations = a chemical equation that doesn’t show the amount of the reactants and products Physical states of substances are represented by:

s = solid

l = liquid

g = gas

aq = aqueous

Ex.) Fe(s) + O2(g) à Fe2O3(s) Catalyst = a substance that speeds up the process, it is written above the arrow (yields sign) Balancing Chemical Equations

Reactants are the parts

Products are the outcome

Balanced equation must have equal quantities of each element or compound on each side of the equation

Mark Cuddy pg. 325-328 Balancing Chemical Equations CON'T
Example:
1 Frame+2 Wheels+1 Handlebar+2 Peddles ==> FW2HP2 = 1 bicycle
The "2's" that were placed before the wheels and peddles are considered COEFFICIENTS used to balance or make the equation even on each side

COEFFICIENTS: small whole numbers that are placed in front of the formulas in an equation in order to balance it.
BALANCED EQUATION: statement in which each side of the equation has the same number of atoms of each element and mass is conserved.
The atoms in the products are the same atoms that were in the reactants, they are just rearranged.

To write a balanced chemical equation...
1) First write the skeleton equation.
2) Use coefficients to balance the equation so that it obeys the law of conservation mass.
Conservation of mass is the fact listed above (Atoms in the products are the same)
Example: When hydrogen and oxygen are mixed, a spark will initiate a rapid reaction to create water.

The following is an example of a balanced equation:

Responsible for the safe handling, treatment, storage, and transportation of hazardous materials

Often work in the field of inspecting, testing, overseeing cleanup work, and determining whether storage facilities are in compliance with regulations

Entry into the field usually requires a degree from a 4 year college with a major in chemistry or other physical science

Other areas of concentration might be industrial hygiene, environmental health, or engineering

Group 2

Grayce Rose pg.330-331
Classifying Reactions:
There are five general types of reactions:

combination

Decomposition

Single Replacement

Double Replacement

Combustion

A combination reaction is a chemical change in which two or more substances react to form a single new substance.
Example: 2Mg + O2-------> 2MgO

(picture of burning magnesium ribbon, ex. of a combination reaction)

Group #3 Pgs. 342-345

Coeditor: Alex Fishback
Kim Kogut
Emily Taylor

Net Ionic Equations pg 342 Kim Kogut

Chemical reactions are summarized in a shorthand way through equations.

However, they don't show the physical states of the reactants and products.

A net ionic equation is an equation for a reaction in a solution that only shows the particles directly involved.

A complete ionic equation is an equation that shows dissolved ionic compounds as free ions.

Pg. 343 Net Ionic Equations (cont.) Emily Taylor

An ion that appears on both sides of the equation and is not directly involved in the reaction is called a spectator ion.

When you rewrite the equation and leave out the spectator ion, you have the net ionic equation, which is an equation for a reaction in a solution that shows the particles that are directly involved in the chemical change.

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

You have to make sure the ionic charge is balanced when balacing net ionic equations.

The nirate ion is the spectator ion in the following equation (equation #1) and the net ionic equation is when the spectator ions are removed (equation #2), making the second equation unbalanced.

Equation #2 is unbalanced because the single unit of positive charge is on the reactant side of the equation and two units of positive charge is on the product side.

By placing a number �428

429� in front of Ag+(aq) balances the charge and placing a number �430

Mass Conservation in Chemical Reactions and Practice problems Pg. 357-358 by: Anne O'Toole

Group 5

Steps in solving a mass mass problem Brendan Lynch
1. Change the mass of G to moles of G (mass G---mol G) by using the molar mass of G.
2. Change the moles of G to moles of W (mol G---mol W) by using the mole ratio from the balnaced equation.
3. Change the moles of W to grams of W (mol W----mass W) by using the molar mass of W.
Any mass given- G Any wanted mass- W

In a typical stoichiometric problem, the given quantity is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires.

Calculating Moles of a Productpg. 364-367 Erika Paiva

Example Problem:

How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolisis according to this balanced equation?
2H2O (l) ----- 2 H2 (g) + O2 (g)

1. Analyze: list the knowns and unknowns
KNOWNS
- mass of water= 29.2 g H2O
- 2 mol H2O = 1 mol O2 (from the balanced equation)
- 1 mol H2O = 18.0 (molar mass)
- 1 mol O2 = 6.02 x 10^23 molecules of H2O
UNKNOWNS
- molecules of oxygen = ? Molecules of O2
The following calculations need to be done:
g H2O-----mol H2O---- mol O2-------molecules O2
The appropriate mole ratio relating ml of O to the mol of H2O from the balanced equation is 1 mole O2/2 mol H2O

2. Calculations: solve for the unkowns
29.2 g H2O x mol H2O/18.0 g H2O x 1 mol O2/ 2 mol H2O x 6.02 x 10^23 molecules of O2/ 1 molecule O2 = 4.88 x 10 ^23 molecules O2.
- For the calculations you start with the Given information, change t mole, mole ratio, and finally change to molecules.

3. Evaluate : Do my result make sense?
The answer should have three significant figures.

Volume-Volume Stoichiometric Calculations
Example Problem

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? 2 NO (g) + O2 (g) ------- 2 NO2 (g)

1. Analyze list the knowns and the unknowns
- volume of oxygen = 34 L O2
- 2 mol NO2/ 1 mol O2 ( mole ratio from the balanced equation
-1 mol O2= 22.4 L O2 (at STP)
- 1 mol NO2 = 22.4 L NO2 (at STP)
Unknown: volume of nitrogen dioxide = ? L NO2

2. Calculate: solve for the unkowns
34 L O2 x I mol O2/22.4 L O2 x 2 mol NO2/ 1 mol O2 x 22.4 L NO2/ 1 mol NO2 = 68 L NO2
- You start with the given information, change to moles, molar ratio, and finally change to liters.

3. Evauate: does the result make sense?
The answer should have to significant figures.

Finding The volume of a gas needed for reaction
Example Problem
Assuming STP, how many milliliters of oxygen are needed to produce 20.4 ml SO3 according to this balanced equation
2 SO2 (g) + O2 (g) ----- 2 SO3 (g)

1. Analyze: list the knowns and unknowns
- volume of sulfur trioxide = 20.4 mL
- 2 mL SO3/1 mLO2 (volume ratio from the balanced equation)
Unknown:
- volume of oxygen = ? ML O2

2. Calculate: solve for the unkowns
20.4 mL SO3 x 1 mLO2/ 2 mL SO3 = 10.2 mL O2

3. Does the result make sense?
The answer should have three significant figures

gunthersclass.com

Group 6

Co Editor: Julia McNamara
Group Members: Julia McNamara, Shannon Degnan, Kelsey Sullivan

Limiting Reagentpg. 368-370 by: Shannon Degnan

In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

an equation could be on a macroscopic scale, which are the interacting particles, or basically just the amount the the particle

an equation could also be on a macroscopic scale, which is the interacting moles

the "limiting reagent" is the reagent that determines the amount of productthat can be formed by a product.

the excess reagent is the reagent that isn't completely used up.

when solving these types of problems, solve it like you are balancing an equation and list your knowns, and unknowns

next, you should find the mole to mass ratio

by using the dimensional analysis we have learned in class, you can find out what the limiting and excess reagents will be

Shannon Degnan

Percent Yield pgs 371-373 By: Julia McNamara

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

Actual Yield – The amount of product that actually forms when a reaction is carried out in the laboratory

Percent Yield – Ratio of the actually yield to the theoretical yield, displayed as a percent

*Percent Yield = (actually yield / theoretical yield) x 100

HOW TO FIND THEORETICAL YIELD:

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

CaCO3-->CaO + CO2

mass of calcium carbonate = 24.8 g 1 mol CaCO3 = 100.1 g 1 mole CaO = 58.1g 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) = 13.9 g aO

Calculating the theoretical yield of a reaction pgs#374-375 By: Kelsey Sullivan

List the knowns and unknowns

Solve for the unknown

Then evaluate, ask yourself if your answer makes sense

Percent yield

multiply ratio of actual yield to theoretical yield by 100%

## Balancing Chemical Equations

Editor: Erin GarrityGroup 1

David O'Brien p. 323-324

Chemical EquationsChemical Equations = a representation of a chemical reaction

- The reactant is on the right, an arrow in the middle pointing towards the products on the left

Ex.) Fe + O2 à Fe2O3Ex.) AgNO3 + H2S à Ag2S + HNO3

Skeleton Equations = a chemical equation that doesn’t show the amount of the reactants and products

Physical states of substances are represented by:

- s = solid
- l = liquid
- g = gas
- aq = aqueous

Ex.) Fe(s) + O2(g) à Fe2O3(s)Catalyst = a substance that speeds up the process, it is written above the arrow (yields sign)

Balancing Chemical EquationsMark Cuddy pg. 325-328

Balancing Chemical Equations CON'TExample:

1

Frame+2Wheels+1Handlebar+2Peddles ==> FW2HP2 = 1 bicycleThe "2's" that were placed before the wheels and peddles are considered COEFFICIENTS used to balance or make the equation even on each side

COEFFICIENTS: small whole numbers that are placed in front of the formulas in an equation in order to balance it.

BALANCED EQUATION: statement in which each side of the equation has the same number of atoms of each element and mass is conserved.

The atoms in the products are the same atoms that were in the reactants, they are just rearranged.

To write a balanced chemical equation...

1) First write the skeleton equation.

2) Use coefficients to balance the equation so that it obeys the law of conservation mass.

Conservation of mass is the fact listed above (Atoms in the products are the same)

Example: When hydrogen and oxygen are mixed, a spark will initiate a rapid reaction to create water.

The following is an example of a balanced equation:

Picture by Mark Cuddy

Lauren Altmeyer pg. 329

Hazardous Materials Specialist## Group 2

Grayce Rose pg.330-331Classifying Reactions:

There are five general types of reactions:

- combination
- Decomposition
- Single Replacement
- Double Replacement
- Combustion

A combination reaction is a chemical change in which two or more substances react to form a single new substance.Example: 2Mg + O2-------> 2MgO

(picture of burning magnesium ribbon, ex. of a combination reaction)

## Group #3 Pgs. 342-345

Coeditor: Alex FishbackKim Kogut

Emily Taylor

Net Ionic Equations pg 342Kim Kogutnet ionic equationis an equation for a reaction in a solution that only shows the particles directly involved.complete ionic equationis an equation that shows dissolved ionic compounds as free ions.Pg. 343 Net Ionic Equations (cont.)Emily Taylor- An ion that appears on both sides of the equation and is not directly involved in the reaction is called a spectator ion.
- When you rewrite the equation and leave out the spectator ion, you have the net ionic equation, which is an equation for a reaction in a solution that shows the particles that are directly involved in the chemical change.

Ag +(aq) + Cl-(aq)--> AgCl(s)- You have to make sure the ionic charge is balanced when balacing net ionic equations.
- The nirate ion is the spectator ion in the following equation (equation #1) and the net ionic equation is when the spectator ions are removed (equation #2), making the second equation unbalanced.

1. Pb(s)+ AgNO3(aq)--> Ag(s)+ Pb(NO3)2(aq)2. Pb

(s)+ Ag+(aq)--> Ag(s)+ Pb2+(aq)- Equation #2 is unbalanced because the single unit of positive charge is on the reactant side of the equation and two units of positive charge is on the product side.
- By placing a number �428

Pb## 429� in front of Ag

+(aq)balances the charge and placing a number �430## 431� in front of Ag

(s)balances the atoms again.(s)+ �432## 433�Ag+

(aq)--> �434## 435�Ag

(s)+ Pb2+(aq)Website: http://www.utc.edu/Faculty/Gretchen-Potts/chemistryhelp/ionic.htmThis website includes steps on how to balance equations and a few practice problems!Predicting the Formation of a Precipitatepgs 344-345 Alex Fishback## - ( refer to chart ).

## (picture by Alex Fischbach)

Co-editor: Anne O'Toole pg. 356-358Group 4-- The Arithmetic of Equations Pg. 353-358Marybeth Nametz pg. 354-355

Lauren Bedard pg. 353-354

Using Everyday EquationsPg. 353-354 by: Lauren BedardUsing Balanced Chemical Equations and Practice problemsPg. 354-355 by: Marybeth NametzChemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reactionquantityof one substance in a reaction, you can calculate the quantity ofany other substanceconsumed or created in the reactionQuantityusually means the amount of a substance as expressed ingrams or molesliters, tons, or moleculesstoichiometric calculationsChemists can track reactants and products in a reaction by stoichiometryIt allows chemists to tally the amount of reactants and products using ratios of moles or representative particlesBy Marybeth Nametz http://www.mikeblaber.org/oldwine/chm1045/notes/Stoich/Equation/Stoich01.htm

(picture of a balanced chemical equation)

Interpreting Chemical Equations Pg. 356 by: Anne O'ToolePicture by; Anne O'Toole http://www.lisisoft.com/imglisi/6/Science/3302methane.gif

Mass Conservation in Chemical Reactions and Practice problems Pg. 357-358 by: Anne O'TooleGroup 5Steps in solving a mass mass problemBrendan Lynch1. Change the mass of G to moles of G (mass G---mol G) by using the molar mass of G.

2. Change the moles of G to moles of W (mol G---mol W) by using the mole ratio from the balnaced equation.

3. Change the moles of W to grams of W (mol W----mass W) by using the molar mass of W.

Any mass given- G Any wanted mass- W

In a typical stoichiometric problem, the given quantity is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires.

(picture by brendan lynch)

http://www.fsj.ualberta.ca/CHIMIE/learning_tools/li4.gif

Calculating Moles of a Productpg. 364-367Erika PaivaExample Problem:

How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolisis according to this balanced equation?

2H2O (l) ----- 2 H2 (g) + O2 (g)

1. Analyze: list the knowns and unknowns

KNOWNS

- mass of water= 29.2 g H2O

- 2 mol H2O = 1 mol O2 (from the balanced equation)

- 1 mol H2O = 18.0 (molar mass)

- 1 mol O2 = 6.02 x 10^23 molecules of H2O

UNKNOWNS

- molecules of oxygen = ? Molecules of O2

The following calculations need to be done:

g H2O-----mol H2O---- mol O2-------molecules O2

The appropriate mole ratio relating ml of O to the mol of H2O from the balanced equation is 1 mole O2/2 mol H2O

2. Calculations: solve for the unkowns

29.2 g H2O x mol H2O/18.0 g H2O x 1 mol O2/ 2 mol H2O x 6.02 x 10^23 molecules of O2/ 1 molecule O2 = 4.88 x 10 ^23 molecules O2.

- For the calculations you start with the Given information, change t mole, mole ratio, and finally change to molecules.

3. Evaluate : Do my result make sense?

The answer should have three significant figures.

calculations

Volume-Volume Stoichiometric Calculations

Example Problem

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? 2 NO (g) + O2 (g) ------- 2 NO2 (g)

1. Analyze list the knowns and the unknowns

- volume of oxygen = 34 L O2

- 2 mol NO2/ 1 mol O2 ( mole ratio from the balanced equation

-1 mol O2= 22.4 L O2 (at STP)

- 1 mol NO2 = 22.4 L NO2 (at STP)

Unknown: volume of nitrogen dioxide = ? L NO2

2. Calculate: solve for the unkowns

34 L O2 x I mol O2/22.4 L O2 x 2 mol NO2/ 1 mol O2 x 22.4 L NO2/ 1 mol NO2 = 68 L NO2

- You start with the given information, change to moles, molar ratio, and finally change to liters.

3. Evauate: does the result make sense?

The answer should have to significant figures.

Finding The volume of a gas needed for reaction

Example Problem

Assuming STP, how many milliliters of oxygen are needed to produce 20.4 ml SO3 according to this balanced equation

2 SO2 (g) + O2 (g) ----- 2 SO3 (g)

1. Analyze: list the knowns and unknowns

- volume of sulfur trioxide = 20.4 mL

- 2 mL SO3/1 mLO2 (volume ratio from the balanced equation)

Unknown:

- volume of oxygen = ? ML O2

2. Calculate: solve for the unkowns

20.4 mL SO3 x 1 mLO2/ 2 mL SO3 = 10.2 mL O2

3. Does the result make sense?

The answer should have three significant figures

gunthersclass.com

## Group 6

Co Editor: Julia McNamaraGroup Members: Julia McNamara, Shannon Degnan, Kelsey Sullivan

Limiting Reagentpg. 368-370by: Shannon DegnanShannon Degnan

Percent Yield pgs 371-373By: Julia McNamaraHOW TO FIND THEORETICAL YIELD:

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

CaCO3-->CaO + CO2

mass of calcium carbonate = 24.8 g

1 mol CaCO3 = 100.1 g

1 mole CaO = 58.1g

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)

= 13.9 g aO

Calculating the theoretical yield of a reaction pgs#374-375By: Kelsey SullivanCalculating the percent yield of a reaction