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) (: