Thursday, December 2

Today we started off class by reviewing the 3.4 Reading Sheet. After looking over that, Mr. Henderson passed out calculators and told us how to use them more efficiently. There is a button (sto) that can store a numbers in the calculator. For example, if you wanted to store a molar mass because you would be using it in the problem a lot, you would plug that number under sto 1. So when you used that number, you would not have to type it in; you would just have to the recall 1 button. He also said that we should either be comfortable with his calculators or bring your own to use. If you bring your own calculator it cannot be an inspire.

We then went over what would the format of the test would like like.

-14 multiple choice questions

-3 pages of short answers (stoichiometry problems)

-1 question on the lab

To study for these types of problems there are review questions on the website and moodle with answers. A great way to study would be to use your packet. Everyone should have almost a completely filled out packet with correct answers. You could rework a problem that is difficult for you and check your answer with your packet. To prepare for the test in class, we completed review questions 17-19 which simulated what the questions about the lab would look like.

Furthermore, Mr. Henderson lectured us about percent yield in detail. To find the percent yield of a substance you need the theoretical yield (mass of the expected product) and the actual yield (mass of the measured product). You discover the theoretical yield by using the normal stoichiometry conversion factors, and you discover the actual yield by massing what your results are in the lab. The equation for the percent yield is:

Percent yield = Actual Yield * 100

Theoretical Yield

Lastly, we attempted problems involving this percent yield concept on a worksheet Mr. Henderson passed out. The toughest problem was number 5; the answer to it was 709.75 grams. Today was a very productive class period as we got in needed review time for the test. Finally, I would like to wish everyone good luck!

Tuesday November 30th, 2010

Todays class started out by Mr. H informing us that we are not going to continue the copper lab that we had started yesterday. We then started to review what would be on our chapter 5 that is this Friday. The test on Friday will consist of 15 multiple choice questions, 3 pages of problems in which you have to show your work, and out of those three pages one page is lab based, another one is 4th grade stoichiometry, and the last page has problems from 7th, 8th, 9th, and 10th grade stoichiometry. There will be a test review page up on the gbschemphys website by Thursday. Another great way to study for this would be to go through the Period 3, Unit 5 delicious book marks. Mr. H also let us know that since a majority of the class will be on a field trip tomorrow, we will be in a computer lab working on the web-assign that is due Thursday.

We then turned to page 13 in our packets and started work on question number 23, part b.

In part B it says... If 215 g of C6H6 are combined with 590 g of O2, what is the limiting reactant and what mass of excess reactant will be left over?

They have also given you the balanced chemical equation:

2C6H6 + 15 O2 ----> 12 CO2 + 6H2O

now you can find the mass of CO2 once reacted with C6H6 and O2.

215g of C6H6 x 1 mol C6H6 x 12 mols CO2 x 44.03g/mol of CO2

------------- ------------- -------------------- = 727.62CO2

78.06 g/C6H6 2molC6H6 1 mol CO2

LEVEL 9 STOICHIOMETRY

590g of O2 x 1 mol O2 x 12 mols CO2 x 44.03g/mol of CO2

------------- ------------- --------------------= 648.74CO2

32.02 g/O2 15molC6H6 1 mol CO2

O2 is the limiting reactant.

590 g O2 x 1 molO2 x 2 mol C2H2 x 78 C2H2g

---------- ------------ --------- = 192 g C2H2 left over

32 gO2 15molO2 1mol C2H2

215-192= 23g

We were then told to work on page 12 number 18 and page 14 number 24.

18. 2KClO3 -------->2KCl + 3O2

o.550g KClO3 x 1 mol KClO3 x 3 mol O2

------------ ---------- = 0.00673 mol O2

122.55gKClO3 2 mol KClO3

24. B2O3 + 6 HF --> 2BF3 + 3 H2O

a. 85 g of B2O3 x 1mol B2O3 x 2 BF3 mol x 67.78gBF3

----------- ------------ ----------- = 165.43 BF3

69.65gB2O3 1B2O3mol 1 mol BF3

115g of HF x 1 mol HF x 2 BF3 mol x 67.78gBF3

---------- ----------- ------------ =129.97 BF3

19.99gHF 6HFmol 1 mol BF3

129.97 -- HF is the limiting reactant

115g of HF x 1 mol HF x 1 BaO3 mol x 69.65gBaO3

---------- ----------- ------------ = 66.78 BaO3

19.99gHF 6HFmol 1 mol BaO3

Excess: 18.22

PSYCF= Please Show Your Conversion Factors (not cute frogs) (: