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) (:

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.