Friday November 19th

Today, Mr. H began the class by going over Alex's blog and showing the variety of pictures that were from Wednesday's lab. He then refreshed our memories about the chemical reactions we had done the day before in our packets. (These types of conversions are going to be on tonight's webassign). Mr. H then proceeded to tell us that he had taken each lab group's silver precipitate, and put them in an oven to dry. He mentioned that he would take them out at the end of class so we could find the mass of each beaker that contained pieces of silver.

Mr. H then instructed the class to open our unit 5 Stoichiometry packets to page 6. There, we worked on the second half of the page; it was titled: Decomposition StoiCHEoMISTRY. The class worked on problems 12-15, and then Mr. H showed us the answers so we could correct our mistakes. He went on to answer any questions that students had about these problems. The answers to problems 12-15 are as followed:

12) 2H2O=> 2H2 + O2

13) 364.8 grams O2- this was a 2 step problem ( converting moles of H2O to grams of O2 )

14) 35.67 grams H2- this was a 3 step problem (converting grams of H2O to grams of H2)

15) 1.22 moles O2- this was another 2 step problem (converting grams of H2O to moles of O2)

In order to solve #13, this is how you would show your work:

         22.8 mol H2O * 1 mol O2 * 32g O2= 364.8g O2
                                  2 mol H2O   1 mol O2

Next, Mr. H gave a lecture on "Limiting Reactants." The notes he wrote on the board are as followed:

Limiting reactant- one reactant that produces the least amount of a product

   2 Al(s)             +          3CuCl2              ==>            3Cu +    2AlCl3

0.010 moles                  0.030 mol    =============> 0.03 mol Cu                           
not enough                    too much-precipitate
dissolves                      (excess reactant)
(limiting reactant)==========================> 0.015 molCu
                                                                                                                                                                  

0.030 mol                    0.030 mol ==============> 0.030 mol Cu
too much                      not enough
precipitate                    dissolves
(excess reactant)          (limiting reactant)
|
|
|======================================> 0.045 mol Cu

The reason why you choose the number that is less than the other one is because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H introduced us to the next level of stoicheometry and used the notes he gave us in class to solve problem #12 on page 4. In this problem, you have to convert the number of grams of each element to the moles of Ti3N4. This is how you would solve this problem:

162g Ti * 1 mol Ti * 1 mol Ti3N4 = 1.13 mol Ti3N4  <== correct answer
                47.88g Ti   3 mol Ti

79 g N2* 1 mol N2 * 1 mol Ti3N4 = 1.41 mol Ti3N4
               28.01g N2   2 mol N2 

1.13 mol Ti3N4 is the correct answer, because as stated in the notes: you choose the number that is less than the other one because after you conduct 1 experiment, you'll run out of one of the elements and you won't have enough  of the second element to create another product.

Next, Mr. H showed us 7th grade Stoicheometry. This was on pages 7 and 8. This page included practicing finding limiting ingredients and the number of cakes produced. For each problem, we had to look off of the original equation, which was: 6 cups of flour + 4 cups of sugar + 2 tbsp baking powder + 3 cups of milk ==> 2 cakes, and find by how much each ingredient was being multiplied. By doing so, you could find the limiting reactant ( the lowest number) and then multiply that number by the original number of cakes produced from the given equation. The answers for page 7 are as followed:

a) Limiting ingredient: milk               # of cakes produced: 1

b) Limiting ingredient: baking powder            # of cakes produced: 3

c) Limiting ingredient: flour               # of cakes produced: 10

d) Limiting ingredient: baking powder                          # of cakes produced: 100

e) Limiting ingredient: baking powder           # of cakes produced: 100

On page 8, was more practice that involved finding the limiting reactant and the # of products produced. We only did #3 on this page. The answers for # 3 are as followed:

Molecules before reaction             Type and # of molecules after reaction                           Limiting Reactant
3 H2 + O2============>       2 H2O
                                                      1 H2                                                                                   O2

6 H2 + 3O2===========>       6H20                                                                                  none

18 H2 + 4 O2==========>      8 H2O
                                                     10 H2                                                                               O2

Mr. H ended the class by getting each lab groups' beaker from the oven, and then had 1 member of each lab group go and mass the beaker.The students that missed class that day, should ask one of their group members for that information. **Note, there is a QUIZ  on Monday, so be sure to study.