February 4th, 2010: Energy & Percent Yield (Zac)

-Enthalpy is the energy stored in chemical bonds
-symbol of Enthalpy is H
-Change in enthalpy is ΔH
-In exothermic rxns enthalpy
-In endothermic rxns enthalpy increases










Calorimetry
- To experimentally determine the heat released we need to know 3 things
1. Temperature change (ΔT)
2. Mass (m)
3. Specific Heat Capacity (C)

These are related by the equation:
ΔH=mCΔT

Example:
Calculate the heat required to warm a cup of 400g of water (C=4.19J/g°C) from 20.0°C to 50.0 °C.
Δ  H = (400)(4.18)(30)
Δ  H = 50160 °C

Percent Yield
-the theoretical yield of a reaction is the amount of products that should be formed
-the actual amount depends on the experiment
-the percent yield is like a measure of success (how close is the actual amount to the predicted amount?
% Yield = Actual/theoretical x 100


Example
-In a double replacement reaction Silver Nitrate is reacted with Barium Sulphate.  If 2.0 mol of Silver nitrate react and 468g of silver sulphate are produced, determine the theoretical yield of Ag2SO4, and determine the percent yield.
 
First, balance the equation: 2AgNO3 + BaSO4 = Ag2SO4 + Ba(NO3)2
Then, determine the theoretical yield of Silver sulphate: 2.0 mol x 1/2 x 312g/1mol = 312g of Ag2SO4
then divide actual by theoretical and multiply by 100: 468/312 x 100: 150 % yield

Feb.8/2011:Limiting Reactants (Brian)

In chemical reactions, usually one chemical gets used up before the other
-The chemical used up first is called the limiting reactant
-Once it is used up the reaction stops
-L.R determines the quantity of products formed
-To find the L.R assume one reactant is used up, determine how
  much of this reactant is required

example:
Find the L.R when 2.8mol of H2 reacts with 1.25mol of O2
2H2+O2 -----> 2H2O
2.8 x 1 = 1.4 mol L.R or 1.25mol x 2 = 2.5
         2                                      1

Feb.8/2011:Percent Yield (Brian)

-The theoretical yield of a reaction is the amount of products that should be formed
-The actual amount depends on the experiment
-The percent yield is like a measure of success

* % yield = Actual      multiplied by 100
                Theoretical  example:
in a double replacement reaction silver nitrate is reacted with barium sulphate. If 2.0mol of silver nitrate react and 468g of silver sulphate are produced.
-determine the theoretical yield of Ag2SO4
-determine the percent yield
2AgNO3+BaSO4 ----> Ba(NO3)2+Ag2SO42.0mol x 1 x 311.9g = 311.9g                          468g x 100 = 150%
               2    1mol                                          311.9g

February 2, 2011: Other Conversions, Volume and Heat (Angelo)

Volume and Heat:
- volume  at STP can be found using the conversion factor 22.4 L/mol (litres/mol)
- heat can be included as a separate term in chemical reactions (this is called ENTHALPHY)
- reactions that release heat are exothermic
- reactions that absorb heat are endothermic
- both can be used in stoichiometry

EXAMPLES:


1. If 5.0 grams of Potassium chlorate decompose according to the reaction below, what volume of Oxygen gas (at STP) is produced?

Create your balanced equation: 2KClO3 -----> 2KCl + 3O2
Create your equation: 5.0 grams * (1 mol / 122.5 g.) * (3 mols / 2 mols) * (22.4 L. / 1 mol) = ?
Calculate: 5.0 grams * (1 mol / 122.5 g.) * (3 mols / 2 mols) * (22.4 L. / 1 mol) = 1.4 L.

January 27: Mass to Mass Conversions (Angelo)

Mass to Mass Conversions:
- mass to mass problems involved one additional conversion
- this chart is a visual example of how to perform mass to mass conversions
GRAMS OF A -------> MOLES OF A -------> MOLES OF B -------> GRAMS OF B


EXAMPLES:


1. Lead (IV) Nitrate reacts with 5.0 grams of Potassium Iodide. How many grams of Lead (IV) nitrate are required for a complete reaction?

Create your balanced equation: Pb(NO3)4 + 4KI -----> 4KNO3 + PbI4
Create your equation: 5.0 grams (KI) * (1 mol / 166 g. (KI)) * (1 mol (Pb(NO3)4) / 4 mols (KI)) * (455.2 g. (Pb(NO3)4) / 1 mol) = ?
Calculate: 5.0 grams (KI) * (1 mol / 166 g. (KI)) * (1 mol (Pb(NO3)4) / 4 mols (KI)) * (455.2 g. (Pb(NO3)4) / 1 mol) = 3.4 grams (Pb(NO3)4)

January 25, 2011: Moles to Mass and Mass to Moles (Angelo)

Moles to Mass and Mass to Moles:
- some questions will give you an amount of moles and ask you to determine the mass
- converting moles to mass only requires one additional step

EXAMPLE:


1. How many grams of Bauxite (Al2O3) are required to produce 3.5 mols of pure Aluminum.

Create your balanced equation: 2NO3 -----> 4Al + 3O2
Create your equation: 3.5 mol (Al) * (2 mols (Al2O3) / 4 mols (Al)) * (102 grams (Al2O3) / 1 mol (Al2O3)) = ?
Calculate: 3.5 mol (Al) * (2 mols (Al2O3) / 4 mols (Al)) * (102 grams (Al2O3) / 1 mol (Al2O3)) = 1.8 x 10^2 grams


2. How many moles of Lead (II) Nitrate are consumed when 4.5 grams of Sodium Sulphide completely react?

Create your balanced equation: Pb(NO3)2 + Na2S -----> PbS + 2NaNO3
Create your equation: 4.5 grams * (1 mol / 78.1 g.) * (1 mol / 1 mol) = ?
Calculate: 4.5 grams * (1 mol / 78.1 g.) * (1 mol / 1 mol) = 0.058 grams

January 21, 2011: Mole to Mole Conversions (Angelo)

Mole to Mole Conversions:
- coefficients in balanced equations tell us the number of moles reacted or produced
- they can also be used as conversion factor:
3X + Y = 2Z
- WHAT YOU NEED over WHAT YOU HAVE


EXAMPLES:


1. If 0.20 mol of Methane are consumed in a combustion reaction, how many moles of CO2 are produced?

Create your balanced equation: CH4 + 2O2 -----> CO2 + H2O
Create your equation:  0.20 mol (CH4 ) * (1 mol (CO2) / 1 mol (CH4)) = ?
Calculate: 0.20 mol (CH4 ) * 1 mol (CO2) / (1 mol (CH4) = 0.2 mol (CH4))

2. When 2.0 mol of Copper (II) react with Iron (II) Chloride, how many moles of iron should be producted?

Create your balanced equation: Cu + FeCl2 -----> CuCl2 + Fe
Create your equation: 2.0 mol * (1 mol / 1 mol) = ?
Calculate: 2.0 mol

NOTE: When you are comfortable, you may remove what type of element, compound, etc. you are using.