Friday, December 04, 2009
I am leaving last weeks web up for now so you can try to figure out these problems
For Monday: Lab Reports  All entries on the Calculation table must be supported by labled calculations! In addition do p. 321 # 25. and 26

Tuesday, December 01, 2009
I decided to start fresh.  Here is how you do the homework sheet.
1. Write the balanced equation
2. Below the equation put the recipe that relates
3. Above the equation put the problem
Example:

1.  How many moles of oxygen may be produced from the decomposition of 3.6 moles of potassium chlorate?
                         

 problem                          3.6 mol                      X =5.4 mol           Want = Given

 equation                       2KClO3  -->  2KCl  +  3O2         or    Mol O2 = 3.6 moles KClO3  x  3moles O2
  recipe                              2 mol                         3mol                                     1                         2 mol KClO3
Do the first 5 problems, we will go over the Lab Report Tomorrow
Wednesday, December 02, 2009
For tonight solve the following problems
1.  In the lab you reacted sodium bicarbonate with hydrochloric acid.  If you used 2.00g of sodium bicarbonate what mass of salt should you produce?  How many moles of salt did you produce?  What if you used 4.0 g of sodium bicarbonate how many grams of salt would you produce.  Now try to answer questions 1-8 on your Lab.
2.  When you breathe you take in oxygen to burn glucose (C₆H₁₂O₆) and you produce carbon dioxide and water.
If you burned 2000g of glucose what mass of products would be formed?
3.  What mass of iron would be fromed from the reaction of 0.25 moles of aluminum with 10.0 g of iron III sulfate?
Thursday, December 03, 2009
For tonight work on your lab report that is due on Monday and the above problems.  I will work #3 below.


                                10.0g            .25mol
                            Fe2(SO4)3  +  2Al  -->  Al2(SO4)3  +  2Fe
                                 399.5g         2moles
You need to find which reactant is in excess or which one is the limiter.  The real recipe is 399.5 g of iron III sulfate combines with 2 moles of Al.  So if you had 10.0 g of iron III sulfate how many moles of Al would be consumed?

   Want= Given
Moles Al =  10.0 g iron III sulfate   x    1 mol iron III sulfate    x   2 moles Al
                                 1                                399.5g                            1 mol iron III sulfate     =.05moles of Al 
Since they gave you .25moles of Al and you only need .05moles Al there is an excess of .20moles Al.  The .25 can not be used in the stochiometry because it does not conform to the real recipe. 
               
                      10.0g          .05 moles                      X_________g
                Fe2(SO4)3  +  2Al  -->  Al2(SO4)3  +  2Fe
                       399.5g         2moles                          2(55.8g)
                                                                                 111.6g
There are two ways to solve for X
1.    10.0g x 111.6g / 399.5g = 2.79g
2.   .05moles x 111.6g/2moles =2.79g