Types of Reactions

Remember the 6 different types of reactions we learned when we were in grade 9, 10?
Synthesis, Decomposistion, Single Replacement, Double Replacement, Combustion and Neautralization? Don't worry if you forgot we're gonna review them today!

Synthesis:
      The combination of two or more substances to form a compound
              ex. C + O2 ------>  CO2
              ex. H2 +Cl2 -----> 2HCl
     General Form: A+B----> C


Decomposition:
       Is a reaction being able to break down a molecule into a simpler substance
             ex. 2AgO + heat ----> 4Ag + O2
             ex. 2NO+ heat-----> N2 + O2
     General Formula: AB----> A+B


Single Replacement:
       A reaction involves replacing one atom in a compound by another atom.
             ex. Cu2Cl2 + Fe---> FeCl2 + Cu
             ex. Cl2 + 2KI---> I2 + 2KCL
       General Formula: A+BC---> B+AC


How to predict Single Replacements:

- If the reacting element is a metal then trade places with the metal atom in the compound
- If the reacting element is non-metal then trade places with the non-metal atom in the compound 
Use the Activity Reaction Chart linked below...
Activity Series Chart Link

Lets use an example of a metal : CuI2  + Fe

Remember that the metal switches with the metal, so the Copper (Cu) must switch places with Iron (Fe).
It should look something like this CuI2 + Fe ----> FeI2 + Cu

Let's use anther example of a non-metal: CuI2 + Br2

Remember that the non-metal switches with the non-metal, so Iodine (I) switches with Bromine (Br).
It should look like this CuI2 + Br2 ----> BuBr2 + I2




PRACTICE TIME!!!

1) Cl2 + CaBr2 --->

2) Mg + ZnSO4 --->

3) Al + O2 --->

4) ICl --->

5) Zn +S8 --->

6) Zn + Ni(NO3)2 --->

7) NH3 --->

8) Cl2 + KI --->

9) 3Mg + 2AlCl3 --->

10) 2CrBr3 + 3Zn --->

Answers:

1) Cl2 + CaBr2 ---> Br2 + CaCl2

2) Mg + ZnSO4 ---> Zn + MgSO4

3) 4Al + 3O2 ---> 2Al2O3

4) 2ICl ---> I2 +Cl2

5) 8Zn +S8 ---> 8ZnS

6) Zn + Ni(NO3)2 ---> Ni +Zn(NO3)2

7) 2NH3 ---> N2 +3H2

8) Cl2 + 2KI ---> 2KCl +I2

9) 3Mg + 2AlCl3 ---> 3MgCl2 + 2Al

10) 2CrBr3 + 3Zn ---> 3ZnBr2 + 2Cr

More practices use these links:

http://chemistry.about.com/library/formulabalance.pdf

http://chemistry.about.com/library/formulabalance2.pdf

Answers to the practices

http://chemistry.about.com/library/formulabalanceanswer.pdf

http://chemistry.about.com/library/formulabalance2answers.pdf

STARTING A NEW CHAPTER WITH... BALANCING EQUATIONS

 

WAHOOOOOOOOOOOOOOOOO!

We've all done it, but do you know why?
Recall that mass of REACTANTS in a chemical equation will always equal the mass of PRODUCTS. Mass is neither created or destroyed!

Got that? Yea. You better have >:)

There's not too much to it! All you have to do is make sure that the number of atoms of each kind on the reactant side equal to those on the product side!

E.g. Ethane + Oxygen > Carbon Dioxide + Steam 

C₂H₆ + O₂ > CO₂ + H₂O  

Here are some important pointers!
1) First balance atoms which only occur in one molecular on each side of the equation. In our example, this would be the CARBON and the HYDROGEN

1C₂H₆ + O₂ > 2CO₂ + H₂O 

1C₂H₆ + O₂ > 2CO₂ + 3H₂O

Now we can balance the oxygen on the reactants side according to the number of oxygen atoms in the product.

UH OH. IT DOESN'T WORK. WHY? Because there are an odd number of oxygen atoms on the product side and the O_{2   on the reactant side means an even number of atoms.} To solve this problem, we can double the numbers we have already put in to make 3 an even number while leaving the general proportions unaffected. 

2C₂H₆ + O₂ > 4CO₂ + 6H₂O

So now we have 14 oxygen atoms in total on the product side. 14/2=7 so... 

2C₂H₆ + 14O₂ > 4CO₂ + 6H₂O

TADA! IT BALANCES!
Some additional tips:

Fe₂O₃ + 3 H₂SO₄ >Fe₂(SO₄)₃ + 3 H₂O

-Balance whole groups (e.g. polyatomic or like the SO_{4 above)} whenever possible - try not to separate them into atoms! Notice the 3 in front of the H₂SO_{4, this balances it with the} Fe₂(SO₄)_3.

<sub>-Try to do the balancing in a certain order - don't jump from one atom to another - especially if you haven't finished balancing it.

-Balance atoms in elemental form last (they can often be used to make up the missing one or two atoms without changing the proportions of the entire equation)</sub>

_{-Watch out for atoms like PHOSPHORUS (which appears as P subscript 4) SULPHUR (which appears as S subscript 8) and the HOFBrINCl (Hydrogen, Oxygen, Fluorine, Bromine, Iodine, Nitrogen, Chloride - which all appear as diatomic elements - with a subscript 2)}

Watch this cool video with a great explanation!
And try these nifty examples! >http://www.sciencegeek.net/Chemistry/taters/EquationBalancing.htm

WAPOOOOOOOOH!

Molar Volume of a Gas at STP.

               

-STP, otherwise known as Standard Tempereature and Pressure, is a standard conditon to compare volume of gases

-STP is equal to.. 1 atmosphere of pressure and with a temperature of 0°C or 273.15 K (Kelvins)

-at STP, 1 mole of gas occupies 22.4 L

THEREFORE, new conversion factors : P

22.4 L of gas                  OR                 1 mole of gas

1 mole of gas                                         22.4 L of gas

Basically, you can add these two new conversions to the mole conversions map from before. You can only use the two conversions between mole and gas.

Ex. A cylinder contains 15.85 moles of N₂ gas at STP. What is the volume of the gas?

15.85 moles x 22.4 L of gas     = 335.0 L of gas
                           1 mole

Diluting Solutions to Prepare Workable Solutions

Making solutions of any concentration from a more concentrated source is possible. What does concentration mean? It's the amount of substance that exists in a given volume of solution.

Don't get it? Think of concentration as orange juice! Just that one can of orange juice has a high concentration. But then adding the 3 cans of water will make it diluted making it have a lower concentration.
A concentrated solution has a relatively high concentration meaning it has a large amount of substance dissolved in the solution.   
Dilute solutions have relatively low concentration so very little substance dissolved in the solution. These dilute solutions are usually formed when large amounts of solvent (usually water)  is added to a concentrated solution to produce a lower concentration.
Moles of solutes are constant. The only difference is that there's more water in the less concentrated solution. Therefore...

moles of solute before = moles of solute after

In terms of molarity & volume in L...

M1L1 = M2L2

Subscript 1= before & subscript 2= after.

Ex. What volume of 6.00 M HCl is used to make up 2.00L of 0.125 HCl
Remember the equation from above in bold purple! 
Rewrite the equation to fit with the information you have..so it'd be:
L1= M2 x L2  --> 0.125 M x 2.00 L = 0.0417 L
           M1                     6.00 M

Ex. If  1 drop (0.050 mL)  of 0.20 M NaBr is added to 100.00 mL of water, what's the molarity of NaBr in the resulting solution?
0.20 M x 0.050 mL= 1.0 x 10^-4 M
(100.00 + 0.050)mL
Remember when answering the questions to round to the correct amount of sigfigs!

Kekekekekeke.

Molar Concentration

Welcome back from winter break and guess what? It hasn't been easy and work has been piling up... but that's okay because nothing says a little knowledge of Molar Concentration will cheer you up!

Molar Concentration is able to compare the amount of solute dissolved in a certain volume of solution.
Or in other words Molar Concentration is represented by M and it has the units of "mols/L."
The formula is Molarity = moles of solute (mol)/volume of solution (L)
In other words it's M = mol/L.
You can always rearrange them like... mol = M x L and L = mol/M.

Let's do some exercises, shall we?
Okay..
1) Calculate molar concentration that has a solution of 0.505 moles of CaCl2 in 1.600 L of solution.
2)Calculate the number of grams of sodium hydroxide in 2.30 L of a 0.7 M NaOH solution.
3) How many grams of AlCl3 are contained in 80.0 mL of 1.5 mol/L of AlCl3.
4) Calculate the volume of L of 0.780 M NaCl solution that contains 55.21 g NaCl.

* Remember to convert your mL to L!*

Answers:
1) M = 0.505 moles of CaCl2 / 1.600L
         =  0.316 mol/L
2) 2.30 L 0.7 M = 1.61
    MM of NaOH = 40 g/mol
    0.316 mols of NaOH x 40 g NaOh / 1 mol of NaOH
    = 12.64g of NaOH
3)
4) MM of NaCl = 58.5 g/mol
    55.21 g x 1 mol / 58.5 g
    = 0.944
L solution = 9.44444 moles of NaCl / 0.780
                = 12.108L