# I am sooooo stuck with this!

Discussion in 'Odds & Ends' started by lhstennis12, Jan 24, 2010.

1. ### lhstennis12New User

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How many moles of lithium chloride will be formed by the reaction of chlorine with 0.046 mol of lithium bromide in the following reaction?
2LiBr + Cl2 ----> 2LiCl + Br2

2. ### coyfishHall of Fame

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.046 moles lol

An hour ! what have you been doing

3. ### lhstennis12New User

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Trying to find something that will help me do these problems!

4. ### coyfishHall of Fame

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1st step = balance the equation

2nd step = find the mole ratio between what you are given and what you are looking for. For example in this equation the ratio is 2 mol of LiCl for every 2 moles of LiBr. Thus the ratio is 2/2 or 1.

3rd step - solve

Multiply .046 moles LiBr X 2/2 ratio = .046

The whole equation is a 1 to 1 ratio. That means if you add 4 moles of a reactant you will get 4 moles of the product.

In more complicated problems you will be given molecular weights that you will have to convert to moles.

5. ### VyseSemi-Pro

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Do you no how to do density/conversion factor, etc. That kinda thing? Got my test on it wednesday and I kinda get it but its hard. Sig Figs are easy though so that is good.

6. ### VyseSemi-Pro

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O sorry, I am not on that stuff yet.

7. ### coyfishHall of Fame

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You asking me lol??? I know how to do everything! (med school student) But im going to bed after this verdasco mayhem is over.

8. ### VyseSemi-Pro

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Hopefully I will figure it out.
I have a question however. How much free time to play tennis or do whatever do you have as a med student?

9. ### coyfishHall of Fame

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Well I don't start until this summer. I just like to brag that I got in . Regardless I study like an animal right now. But I do 1 thing for myself everyday. I workout 3 times a week and I usually do cardio a couple times. So I might play tennis 1 time during the week and on weekends.

10. ### VyseSemi-Pro

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I am thinking of doing pharmacy. Just curious on your amount of free time. I am doing pre=pharmacy stuff at the moment. This first year hasn't been bad. Plenty of studying but enough time to play tennis/workout/relax, etc. I could always switch at this point, but I think pharmacy could be great.

11. ### coyfishHall of Fame

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Well pharmacy is quite a bit easier than medicine so you will definately have adequate free time. My ex-girflriend actually is going to pharm school. You make good money but be advised that most pharmacists end up counting pills lol . . .

12. ### lhstennis12New User

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Got another:

2A + 7B -----> 4C + 3D

The above is a hypothetical reaction.

Calculate the percentage yield in each of the following cases.

The reaction of 0.0251 mol of A produces 0.0349 mol of C.

Do I do 0.0251/2 to start?

13. ### coyfishHall of Fame

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I assume that question is just a part of a larger question???

Assumming you have given all the info this is how I would do it:

These are a little tricky to explain. You really have to visualize whats going on.

Step 1 is balance which is done
Step 2 is to get the product / reactant ratio: In this problem every 2 mol of A will equate to 4 mol of C in the product. Or 1:2. So the way you wanted to start was actually backwards.

% yield is theoretical / actual

To find theoretical you want to MULTIPLY by the ratio. One way I remember it is that what you are looking for (solving for) is always on top. So you multiply by 2/1 instead of 1/2.

So theoretical = .0251 X 2/1 = .0502

The actual yield .0349

% yield = .0349 / .0502 X 100 = 69.5%

Hope that helped. Its kind of tricky to visualize why you multiply by 2/1 instead of by 1/2.

14. ### KobbleHall of Fame

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http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/limitr15.swf

Review this animation until it makes sense. Essentially, 7B is your limiting reagent. If it (actual moles of 7B) was in excess of the theoretical moles of 7B, you would have twice as many actual moles of C than A. All about the ratio.

1 (A) + 2 (B) = 2 (C) Original balanced equation

2 (A) + 2 (B) = 2 (C) A is in excess.

0.5 (A) + 2 (B) = 1 (C) B is in excess. In this case, A is contributing half the molecules typically needed to complete the yield (formation) of two moles of C. (A) will only react with twice as much (B). See what happened. "A" was reduced by half, yet "C" was still double of "A."

2(A) + 1(B) = 1 (C) A is in excess, again. This is what is similar to what is happening in your problem. In this example, all of "B" is only reacting with 0.5 moles of "A".

Overall, in my example, "C" can never be greater than than twice "A", and "C" can never be greater than "B."

In your problem, "C" can never be greater than twice "A". And if "A" were the limiting reagent, it would mean that "B" is excess, and "C" would still be twice "A."

Forgive me if I made any errors.