Today, we did more practice in stoiceometry with limiting factors. We had the equation "Al + CuCl(2) --> AlCl(3) + Cu". First we had to balance the equation, so we got "2Al + 3CuCl(2) --> 2AlCl(3) + 3Cu". Now that we had the mole ratio, we could figure out which is our limiting reactant with the information of .82g Al and .87g CuCl(2).

To find out which one is the limiting reactant, there are two ways.
The first way, you could find out how much of one reactant you need to react with the other. We used Al for this one. So we take the mass, .82g Al, multiply it by 1 mole over it's molar mass, then find the mole ratio of Al and CuCl(2) (which is 2 to 3), then multiply it by the molar mass of CuCl(2). Since the answer (6.1g CuCl(2) was more than what we have (.87g CuCl(2)), we can conclude that CuCl(2) is our limiting reactant. If out answer turned out to be less than .87g, Al would be our limiting reactant.

The other method is to find out how much of each reactant you need to produce one product. It's always good to look ahead since this question 99.2% of the time will be right after the first: how much of a product you could make. You would follow all of the above steps for each of the reactants and you would choose a product. We used Cu as our finishing product. Remember, whichever reactant produces the least of a product is the limiting reactant. In the end, CuCl(2) produced the least Cu, which turned out to be .41g. Since that's the most that CuCl(2) can produce, that is all that will be produced. You don't do anything with that number!

So after we react this all, how much Al is left? well... to find that there are two ways you could do it, but both are pretty much the same.
First you could take how much of the limiting reactant you have and find out how much of the excess reactant you need to react it, which then you subtract that from how much you start with. To get this, just follow the steps listed in paragraph 2, except for the limiting reactant.

The other way is to take that .41g Cu and see how much of that excess reactant (Al) you need to react with that. It's basically all the same as listed above with different variables.

After we got through all of that, we got to our worksheet, Limiting Reactants, which is part of our homework.
For tomorrow, We need to have our Lab which we recently did and our chemistry question set #1.
1 silver troy ounce = $27.46, I got part of your homework done.
If you're still reading, please come up to me tomorrow and give me a waffle.
We will most likely have a quiz tomorrow, we were lucky not to have one today.


The next scribe will be... Austin W.

Beginning of the End

We ended this week by finishing up the Copper and Silver Nitrate Lab. That is, most of us did. Some people who accidentally dumped their silver before decanting it had to start the whole lab over again. The rest of us used our decanted and dried silver and weighed it. Now that we were all finished with the lab procedure, we got much time to just work on the post lab.

We also had a quiz, although some people haven't taken it yet, so I don't think that I should post the answers. Sorry! Although if you did take the quiz, Mr. Lieberman was super speedy and graded them all before the bell even rang.

The post lab of the Copper and Silver Nitrate lab is due on Tuesday, which is also the last day of the week! Yay!

For other homework, we have to finish the Intro To Limiting Reactions worksheet. If you lost it or misplaced it, you can find it here. If you finished it, you can also find/check the answers here.

We also have the first set of book problems due on Tuesday. The problem numbers are #61-68 on page 70-71. If you want to get ahead, you can do the second set of problems as well, which are due on Wednesday, December first. They are #70-78 on page 71.

Good luck to you guys, and keep working hard. The week's almost over already, and it isn't even Monday.

The next scribe will be Alex K.