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) (:

Tuesday 23, 2010

Today we started out class by turning in our lab notebooks but, it quickly turned into a very interesting start. Mr. Hendersons computer was broken therefore projecting a magenta color. Some people claimed it was a "purpendicular" color crayola made so we went on a search to find that color. Sadly we couldnt find any proof that color existed. After that we went on the wikipedia page for green, then Mr. H asked girl students and boy students what color they see, guys saw black and girls saw a dark shade of green, therefore when were all married, the men just agree to the womens color likings. Also, a different chemistry teacher came in and informed us that only male turkeys gobble. That was a great start to a chemistry class.

After our great begining we turned to page 8, to review 7th grade stoichiometry in a very picturesque way. We only did number 4. *NOTE* the white circles that are just floating around[theres four] should have a diagonal line going through them so you know which ones carbon, same in your answer key. In the after reaction box we drew in two H2O four H2 and four CO. The limiting reactant was Carbon.

After that we turned into page 9 in our packets. The page was fairly easy as we breezed through it. We skipped question 6 going on to seven which was finding the molar mass. Then we went on to questions 8 and 9. *NOTE*= The limiting reactant is the reactant that produces the least amount of product.In question eight we had to figure out which was the limiting reactant and then the max mass of NH3. In nine it was pretty much the same except we had to list the limiting reactant.

Here are the answers to page 9:

8) A)1 mole, 3 moles, 34 grams/ 2 moles.

B) 2 mole, 6 moles, 68 grams/ 4 moles

C) 5 mole, 15 moles, 170 grams/10 moles

9)A) 2.21 moles, 6 moles, H2, 68 grams./4 moles

B) 1 mole, 4 moles, N2, 34 grams/ 2 moles

On to page ten, on page ten it was mostly just practice of 8th grade stoichiometry. We had a balanced equation and we had to find the limiting reactant from two masses we were given.

Here is all the work for page 10:

HOMEWORK!:unit five delicious assignment due Tuesday! It will be graded!

Two webassigns if you want extra credit are due Saturday and Monday.

Also, test Friday and lab notebooks will be collected again!

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.

Thursday, November 18

This morning, we walked into class and were notified that there was a lab today. The lab is called ST1 Copper and Silver Nitrate Lab. The purpose is to use careful measurements to determine the mole ratio of copper(II) and silver in the single replacement reaction of silver nitrate with copper. We were instructed to take out our lab notebooks and copy down this purpose onto a clean right-side page. On the left side, Mr. H told us to write the chemical equation for the reaction, for it would be helpful later. It is:

Cu (s) + 2AgNO3 (aq) --> 2 Ag (s) + Cu(NO3)2 (aq)

We all then received a piece of paper describing the lab's procedure. 2 people from the lab group were to carry out the first procedure while the remaining 2 were to carry out the second procedure. The procedures are as followed:

1) Prepare the Silver Nitrate for Reaction:

• Mass an empty 50 mL beaker. Record.
• Add ~1.4-1.6 g of AgNO3.
• Mass beaker plus AgNO3. Record.
• Add 30 mL of water.
• Stir to dissolve solid; rinse stirring rod so as to avoid loss of the AgNO3.

2) Prepare Copper for Reaction and React:

• Obtain ~25 cm of copper wire.
  
• Mass wire. Record.
  
• Loosely coil the wire so as to rest of a wood splint as shown. It should be long enough to hang on the wood splint and dangle in the solution. Yet it should not touch the bottom of the beaker.
  
• Add 3 drops of nitric acid to the solution; avoid contacting the copper wire with the HNO3. Do not stir.

After we did what was required, we were instructed to leave our notebooks at our lab benches and come back to our seats. This time was set aside for the reaction to occur. When we were back at our desks, Mr. H told us to take out our packets and turn to page 6 for

E V E N M O R E S t o i C H E o M I S T R Y!!!!

Mr. H told us to do number 7; write the balanced equation for the synthesis of magnesium oxide from its two elements - magnesium and oxygen. The answer is as follows:

2Mg (s) + O2 (g) = 2 MgO (s)

Then we were told to do number 8; calculate the molar mass of each of the three reactants and products. The answer is as follows:

Mg: 24.305 g/mol O2: 32.0 g/mol MgO: 40.3 g/mol

After, as a class, we went through the next three problems and identified how many steps would be needed (using the Mole Island method, of course). Number 9 was a one step equation. Number 10 was a two step equation. And number 11 was a three step equation.

The answer to number 9 (Determine the number of moles of magnesium oxide produced by the reaction of 3.25 moles of magnesium.) is 3.25 molecules of MgO, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide.

The answer to number 10 (Determine the mass of magnesium oxide produced by the reaction of 5.22 moles of magnesium.) is 210.39 g MgO, again, simply found by using Mole Island to convert from moles of magnesium to moles of magnesium oxide to grams of magnesium oxide.

The answer to number 11 (Determine the mass of oxygen gas which will react with 65.2 grams of magnesium.) is 42. 92 g O2, aaaaaaand again, simply found by using Mole Island to convert from grams of magnesium to moles of magnesium to moles of oxygen to grams of oxygen.

Before we knew it, it was time to check back at our lab. Our whole lab group was to do the third and fourth procedures listed on the sheet of paper:

3) Retrieve Silver Product from Copper Wire:

• Obtain a clean, dry 100 mL beaker. Using a permanent marker, label it with your period and your names.
• Mass the empty beaker. Record.
• Carefully lift the copper wire (with silver attached) from the beaker.
• Hold wire over and into 100 mL beaker and rinse with a forceful stream of DI water. The goal is to knock the silver off the copper and into the beaker.
• Continue rinsing until all the silver crystals are removed.
• Set labeled beaker aside to dry. Once dry (the next class period), mass the beaker with the silver. Record.

4) Rinse and Mass Unreacted Copper:

• Rinse the remains of the copper wire in an acetone bath.
• Being careful not to break or crumble the copper, thoroughly dry it by dabbing it with a paper towel.
• Mass the remaining copper wire. Record.

We did just that. The untouched product (silver still connected in the original beaker) should look like this:The removal of the silver should look like this:

Mr. H then told us to clean up and reminded us that we would be finishing the lab tomorrow.

The homework is the Stoich WebAssign 2.

Wednesday, November 17, 2010

Hello Class!

Today was a short but fun-filled day. Late arrival means the class was only 35 minutes long. The first thing we did was reviewing our stoichiometry connected to Mole Island. Mole Island is an interesting topic which has to do with knowing how many "division bars" you have in a conversion equation. (For example, if you're converting moles to moles, you multiply once. If you're converting moles to grams, you do it twice. Finally, if you're converting grams to grams, you do it three times.) See pages 3 to 4 in packet for more info. For trickier problems, such as finding out how many kilograms in a certain substance there are, (like on our WebAssign), just do it regularly but when you're done multiply by 1000. (kilo=1000)

But wait, that's not all!

Then, we started stoiCHEoMISTRY. (You get it? It's like chemistry, but inside another word containing other letters) Basically. . . . . same thing. You look at a reaction, balance it, and then you go on and determine molar mass of every molecule. (By using the PT) Then you go on and find how many grams or moles of a particular substance there is in another substance. Again, see pages 5 to 6.

We ended the class by taking a pop quiz. If you have any questions, see Mr. H or simply refer to his webpage. Also, make sure to continue being proactive when it comes to those WebAssigns.

Bye.