Wednesday, February 9, 2011, David Colston

1)First, we picked up two new worksheets. The page numbers were 33 and 34, and they had the same title: Limiting Reactants 4

2) We then proceeded to go over the Penny lab we did in class. In the Penny Lab, we put the CuCl2 and the nails in water and stirred it up. After stirring it for a while, the blue-tint to the water caused by the Copper Chloride disintegrated, and the water remained there. By this we know that CuCl2 is the limiting reactant, but you can also do this by writing the problem out like this:

First you must balance the equation, so it would be:

3CuCl2 + 2Al ----> 2AlCl3 + 3Cu

Then, you need to put your data in your tables:

3.0 grams CuCl2	1 Mole CuCl2	3 Mole Cu	63.55 grams Cu
	134.45 grams CuCl2	3 Mole CuCl2	1 Mole Cu

Answer: 1.417 grams Cu - We know this is the limiting reactant because CuCl2 has less then the Excess reactant, Al

1.25 grams Al	1 Mole Al	3 Mole Cu	63.55 grams Cu
	26.98 grams Al	2 Mole Al	1 Mole Cu

Answer: 4.416 grams Cu - Excess Reactant... Can only make as much as the limiting reactant though, so it can only make about 1.4 grams of CuCl2

To answer the second question, we can go backwards from grams needed to grams CuCl2.

3) We went over quiz number 4

A.

12.5 grams Al	1 Mole Al	2 Mole Al2O3	101.96 grams Al2O3
	26.98 grams Al	4 Mole Al	1 Mole Al2O3

12.5 grams O2	1 Mole O2	2 Mole Al2O3	101.96 grams Al2O3
	32.00 grams O2	3 Mole O2	1 Mole Al2O3

Answers

A. 23.6 grams Al2O3 = Limiting Reactant

and

26.6 grams Al2O3 = Excess Reactant

B. The mass of the product can only be as much as its limiting reactant, so it would equal 23.6 grams

4) We went over a problem on page 33

Problem Number 2: 2NH3 +5F2 ----> N2F4 +6HF

If 120 g of NH3 is allowed to react with 590 g of F2,

What mass of HF can be formed?

120 grams NH3	1 Mole NH3	6 Mole HF	2 grams HF
	17.04 grams NH3	2 Mole NH3	1 Mole HF

Answer =422.7 grams HF or 420 grams HF (Sig Figs)

590 grams F2	1 Mole F2	6 Mole HF	20.01 grams HF
	38 grams F2	5 Mole F2	1 Mole HF

Answer= 372.8 grams HF or 370 grams HF (Sig Figs)

Limiting Reactant

What is the Excess Reactant?

372.8 grams HF	1 Mole HF	2 Mole NH3	17.04 grams NH3
	20.01 grams HF	6 Mole HF	1 Mole NH3

120 grams NH3 -105.8 grams NH3=14.2 grams NH3 or 14 grams NH3 (Sig figs)

5) We took Quiz Number 5, covering the things that we learned in class today (That are on here)

6) We went over the newest Lab, that is just one day out of our 3 day lab test. If you missed yesterday, you may want to see Mr. Tucker, but it isn't the end of the world seeing as we only spent about 10 minutes on it. What we began to do was: We put the CuCl2 and the nails in water and stirred it up. The nails become a rusty gray color. We will work more on this in class tomorrow.

HOMEWORK/NOTES: PREPARE FOR QUIZ, AND FUTURE QUIZZES. IF YOU DON'T UNDERSTAND SOMETHING, GO OVER IT OR GET HELP WITH MR. TUCKER, A FELLOW CLASSMATE OR IN THE TLC. DO AS MUCH OF PAGE 33 and 34 AS POSSIBLE. THINK ABOUT THE LAB EXAM!!!

If you read this, thank you! I hope it was a help!

Tuesday-02/08-Tucker

Today we did a chemistry scene investigation lab to help us understand limiting reactants. It looks like this:

1.The chemical equation for this lab was:

3CuCl2 + 2Al --> 2AlCl3 + 3Cu

2. We worked with 1.25g of aluminum, 3g of CuCl2, and 175mL of water

3. To complete the lab we had to figure out how many chemicals w

e need to produce 1.8g of copper

4. In the lab some of the CuCl2 was used up, some of the aluminium turned into copper but there was still some aluminum left

5. CuCl2 was the limiting reactant

This is a picture of the CuCl2 and pieces of aluminum from the lab:

Homework:

Answer the 2 questions from the lab:

1.Which supplier made the mistake?

2. How much more of the chemicals will you need?

M-2/07-Scott

-Passed back the Mole Quiz. You have untill the end of tomarrow to make it up.

-went over Stochmerty Quiz 3

- Did limiting reactents notes

1) put the two muesurement in grams of the chemical that is given on the left side, with one on top, and another below that (two diffrent equations)

2) Put the the grams of the product and the right side of both equations (jusst like how we set it up before)

3) solve using the mole island going from grams of A, to grams of b.

4) Look at the equation that comes out with the least amount of your product. The intial chemical that was given in grams is your limiting reactent. The other is called the Exessive Reactent.

5) The lowest answer is the most of this product that can be made.

-Took Stochmetry Quiz 4

- Homework: journal page 29

Fri 6/2-Koh

Today we learned how to do limiting reactions.
1. Mr. Tucker showed us how to do Limiting Reactants using the grilled cheese example.
Say you have ten grams of bread and twenty five grams of cheese. How much sandwich would you be able to make? Since ten grams of bread only makes .8 sandwiches (using conversions factors), and since twenty five grams of cheese makes and entire 1.85 sandwiches, (also using simple conversion factors), then we'd only be able to make .8 of sandwiches. We only have enough bread to make .8 sandwiches and the entire 25 grams of cheese would be too much and we'd have excess cheese. This works the same way with molecules. You can't make 2 water molecules with four hydrogen atoms and one oxygen atom. You'd just have one molecule of water and two left-over hydrogen atoms.

2. We went over a limiting reactants sheet. It looked something like this.

3. Lastly we got our quizzes back and took the new one.

Extra Stoich Practice!!

For a problem that gives moles of Substance A and asks for mols of substance B use 1 step.  For this one step you will just use the balanced equation ratios.

For a problem that gives moles of Substance A and asks for grams of substance B use 2 steps.  Start with the balanced equation ratios for the first step and then use the molar mass (from periodic table in grams/mole) of substance B for the second.

For a problem that gives grams of Substance A and asks for mols of substance B use 2 steps.  Start with the moalr mass of substance A and then use the bal. eq. ratios.

For a problem that gives grams of Substance A and asks for grams of substance B use 3 steps.  Start with the molar mass of substance A, then use bal. eq. ratios, and then finish off with the molar mass of substance B.

For a problem that gives ml of Substance A and asks for grams of substance B use 4 steps.  Start with the DENSITY, then use moalr mass of substance A, then bal. eq. raitos, then finish off with the molar mass of substance B.  If you are ever given the grams and asked for ml, just reverse this order.

Now try to identify how many steps the following problems contain and what the steps involve (no need for any math, X and Z just hypothetical substances, just set problems up!!)

1.  How many moles of X react wirh 25.0g of Z?

2.  How many grams of Z will be produced from 12.35g of X?

3.  How many ml of Z can be produced if we also produced 35.245g of X?

4.  How many moles of Z can react with 500.0moles of X?

5.  What mass of Z would be needed to produce 10.1moles of X?

BONUS:  How many atoms of X can be produced from 25g of Z?  Use mol island for this one

Friday 1.28.11

1. Went over Elizabeth's post

2. Went over Sandwich Stoich, numbers 7-10
Number 7

25 Sa(Sandwich)	12.5 g/Br
	1 Sa

=31.25 g/Br
Number 8

25.0 g/Ch	1Sa
	13.5 g/Ch

=1.85 Sa
Number 9

10 g/Ch	1 Sa	12.5 g/Br
	13.5 g/Ch	1Sa

=9 g/Br
Number 10

10 g/Br	1 Br	1 Sa
	6.25 Br	2 Br

=.8 Sa
3. Got 2 pages of notes, but only did the first page on Stoichiometry

4. Learned how to get from Mole A to Mole B by balancing equations (See notes)

5.Journal page 15 was homework

6. Quiz on Monday.