Acids and Bases

Hello Everyone! Happy Tuesday!

So we did a few things in class today, most of which have to deal with the things we have been doing in class (Titration and Titration labs).

1. We went over the homework from last night (Monday) which was the Titration lab #2. Many people may have finished in class, but those of you that didn't had to finish it for homework. If you were absent, make sure you get Mr. Tucker to stamp it in for you. Here are some of the answers we got when going over the lab:

The Molarity of the Base for: Trial 1 is .89 M

Trial 2 is .87 M

Trial 3 is .87 M

Trial 4 is .63 M

Trial 5 is .94 M

1. The known substance is the Molarity Acid (.4 M) and the unknown is the Molarity of the base.

2. You could make the argument that both trial 4 and 5 are deviant trials because they are not so similar to the original three. You could just as easily make the argument that they are close enough, and that they are not deviant trials.

3. (Molarity of base at top)

4. a) .84

b) Again, this depends on which side you argued if you said 4 and 5 were deviants, your average is approximately .877 or .88. If you said they were not deviants, it stays the same at .84M

5. Yes, you should throw out the deviant trials, but you should be sure to mention them in your report so that you do not make that mistake again, and just so you remember there was one or multiple with data that was far off.

2. We went over the beginning of Titration lab 3

3. We went back to our tables in our lab groups, put our safety goggles on, and began our labs.

Basically what we did was we did the same thing we had been doing, but with solid acid opposed to liquid acid, and measuring how much base was needed to turn a set amount of acid pink.

***Homework***

Finish lab questions and anything else you might not have finished in class

Other things to take note of:

Someone took Mr. Tucker's keys, and if you know anything about them, he would appreciate it very much if you let him know.

Also, If you want to sign up for the "Chemistry Challenge" there are still teams that are needed. So, if you want to participate in this, and get a 20/20 for a lab grade, feel free to sign up.

Mr. Tucker has been assuring us that it will be somewhere on the scale of ok to fun, so if you like this kind of thing, sign up!

Hope this is helpful!

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!