Balancing Chemical Equations

Editor: Erin Garrity

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

Rad-reaction_900.gif
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:
external image balanced-equation.gif
Picture by Mark Cuddy

Lauren Altmeyer pg. 329
Hazardous Materials Specialist
  • Keep the environment safe from harmful substances
  • 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
hazmat.jpg

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
external image Magnesium_burning.jpg
(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.
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
    429� in front of Ag+(aq) balances the charge and placing a number �430

    431� in front of Ag(s) balances the atoms again.
Pb(s) + �432
433�Ag+(aq) --> �434
435�Ag(s) + Pb2+(aq)

external image solubilityrules.gif

Website: http://www.utc.edu/Faculty/Gretchen-Potts/chemistryhelp/ionic.htm

This website includes steps on how to balance equations and a few practice problems!


Predicting the Formation of a Precipitate
pgs 344-345 Alex Fishback
  • Some combinations produce precipitates, others do not.
  • Whether or not a precipitate forms, depends upon the solubility of the new compounds that form.
  • You can predict the formation of precipitate by using the general rules for solubility of ionic compounds
- ( refer to chart ).

external image SolTab.gif

(picture by Alex Fischbach)

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

Using Everyday Equations
Pg. 353-354 by: Lauren Bedard


Using Balanced Chemical Equations and Practice problems
Pg. 354-355 by: Marybeth Nametz
  • A balanced chemical equation tells you what amounts of reactants to mix and what amounts of product 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 you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction
  • Quantity usually means the amount of a substance as expressed in grams or moles
  • Quantity could also be expressed in liters, tons, or molecules
  • Stoichiometry is the calculation of quantities in chemical reactions
  • Calculations using balanced equations are called stoichiometric calculations
  • For chemists stoichiometry is a form of bookkeeping
  • Chemists can track reactants and products in a reaction by stoichiometry
  • It allows chemists to tally the amount of reactants and products using ratios of moles or representative particles
external image coeff.gif
By 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'Toole
  • a balanced equation provides the same kind of quantitative information that a recipe does
  • the ingrediants are the reactants and the cookies are the products
  • a balanced chemical equation can be intrepreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume

external image 3302methane.gif
Picture 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'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.


external image li4.gif(picture by brendan lynch)
http://www.fsj.ualberta.ca/CHIMIE/learning_tools/li4.gif

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.

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





external image stoichprobs.jpg
gunthersclass.com

Group 6

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

Limiting Reagent pg. 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

problem-solving-limiting-reagent.jpg
lr-3.jpg
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%
external image Percent-Yield1.jpg
Calculating the percent yield of a reaction
  • List the knowns and unknowns
  • solve by using the percent yield
  • Evaluate whether your answer makes sense