Tuesday, February 22, 2011


The Agenda
  • Stamp the homework (4-6 in Reaction solubility and the lab)
  • Check homework
  • Start Molarity
The answers in 4-6 Reaction Solubility:

       4.) H2CO3(aq)  + 2KOH(aq)         ---> 2H2O + K2CO3(s)
       5.) CdBr2(aq) + Na2S(aq)           ---> CdS(s) + 2NaBr(aq)
       6.) 2NaOH(aq) +Ba(NO3)2(aq)  ---> 2NaNO3(aq) + Ba(OH)2(aq)


  M =moles of solute
         liters of solution

Its like stoich but there's Liters in it. There are some stuff that Mrs. Mandarino has not covered yet,
she just talked about the equation of Molarity and answered some of the questions in Molarity sheet 1 and 2

Molarity sheet 1

1.)What is the molar mass of the following compounds? Write dissociation reactions for each of them.
 (so basically, you just have to find the molar mass- easy)

a.) NaNO3 (1 x 14.01g) + (1 x 22.99g) + (3 x 16.00g) = 85g
b.) CaCl2 (1 x 40.08) + (2 x 35.45g) = 110.48g

2.)Calculate the number of moles of the following (yeah, step by step)

a.) 25g of NaNO3 x 1mol NaNO3 = 294 mol NaNO3
b.) 200g of CaCl2 x    1mol CaCl2   =1.8 mol CaCl2
                                110.98g CaCl2 

3.) Calculate the molarity of the following solutions. (You will use the equation in here- new stuff)

  a.) 0.25 liters of solution contains 0.0346 moles of NaNO3
            M= 0.0346 mol = .1384 mol/L or .138 M
                     0.250 L 
  b.) 7.54 liters of solution that contains 0.75 moles of CaCl2 

            M= 0.75 mol = 15 M
                     7.54 L 

  c.) 100.0 ml of solution that contains 1.5 moles of C6H12O6 (Pro tip: watch out for the measurements, we
       need liters not milliliters, so it's .1L of solution. Also C6H12O6 is Glucose)

            M= 1.5 mol = 15 M
                     .1 L
   d.) 560 ml (.56L) of solution that contains 0.025 moles of AgNO 

            M= .02 mol = .036 M
                    .56 L

Molarity sheet 2

4.) Calculate the molarity of the following solutions.

 a.) 1.5 liters of a solution that contains 55.5 g of NaNO
        55.5g  NaNO3 x 1 mol NaNO3  = .653 mol NaNO3  (because we want the     = .44 M
                                   85.0g NaNO3         1.50L               molarity of the solution)

5.) Calculate the moles of solute in the following solution.

 a.) 0.2 liters of 0.125 M C6H12O6
       .2 L x .125 mol  = .025 mol C6H12O6

6.) Calculate the volume of the following solutions.

 a.) 0.5 moles of CaCl2 makes of a 0.667 M solution

       0.5 mol x        L     =.750 L CaCl2 

Well that's all folks. Remember to finish the other questions in Molarity sheet 1 and 2 for homework tomorrow. Furthermore, the webassign are posted and ready to be answered.

Friday, February 18, 2011


Since Dan has failed us twice, I am sharing 8th period's scribe:

Yesterday in class we finished our Mixtures Lab. We took each of the four mixtures and completed the evaporation test and went on to the filtrate lab. To fill out the back sheet of the paper, we can use the notes we took in class on Wednesday. If you want to check if you got the correct results, they are on moodle.
We also learned about solubility and dissolving. We learned that there are two categories of solutions: the solutes and the solvents.

-What is being dissolved
-Usually smaller amount

-What deos the dissolving
-Usually the greater amount

Ionic Compounds: When they are dissolved in water they split into ions (dissociation)

Covalent Compounds: When they are dissolved in water they do not split

*When the compounds are dissolved in water you change the subscrips to (aq), which means in water*

The homework is to finish the Mixtures Lab.

[The notes Mrs. Mandarino gave us in class are on moodle, in the notes section.
If you missed ChemDay, the make up assignment can be found in the homework log section on moodle.]

Tuesday, February 8, 2011

Limiting Reactants

 Limiting Reactant Strategy

If a problem mentions amounts of more than one reactant, its probably a limiting reactant problem.

  1. Go from each reactant to amount of product (grams of each reactant to grams of product)
  2. The reactant that produces the lower amount of predicted product is the limiting reactant.  This amount of product is also the amount that can be produced.
  3. The other reactant is your excess reactant.

For Example:

Fe + S  à  FeS

If I have 28g of Iron (56 g/mol) and 24g of Sulfur (32 g/mol), how much FeS can be made and what is the limiting reactant?

28 g Fe  x  1 mol Fe  x  1 mol FeS  x  87.92g FeS  =  44 g FeS
                 55.85 g Fe     1 mol Fe       1 mol FeS

24 g S  x  1 mol S  x  1 mol FeS  x  87.92g FeS  =  66 g FeS
                32.07 g S     1 mol S         1 mol FeS

Fe is the Limiting reactant because it runs out first and produces less FeS.

Tuesday, February 1, 2011


Today in Mando's fifth period chem class we did a multitude of things. we went over the homework which was pages fifteen and sixteen. After that we spent the rest of the class taking our Stoich quiz, you must score a one-hundred percent on at least one of the Stoich quizes to take the unit test. the homework for tonight is pages 17 and 18.

Also see the cheat sheet in the Unit 8 Stoich notes on Moodle!!! 

Jack H.