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#1 |
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New User
Join Date: Jun 2009
Posts: 12
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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 Please help, I've tried for an hour.
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Prince Exo3 Grahpite 100s SW forehand, 2HBH, mostly flat and kick serves. |
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#2 |
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Hall Of Fame
Join Date: Feb 2009
Posts: 1,670
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.046 moles lol
An hour ! what have you been doing |
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#3 |
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New User
Join Date: Jun 2009
Posts: 12
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Trying to find something that will help me do these problems!
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#4 |
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Join Date: Feb 2009
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Well the one you asked about is very simple.
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. |
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#5 |
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Semi-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.
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Many of life’s failures are people who did not realize how close they were to success when they gave up. |
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#6 |
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Semi-Pro
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O sorry, I am not on that stuff yet.
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Many of life’s failures are people who did not realize how close they were to success when they gave up. |
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#7 |
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Hall Of Fame
Join Date: Feb 2009
Posts: 1,670
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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.
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#8 | |
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Semi-Pro
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Quote:
I have a question however. How much free time to play tennis or do whatever do you have as a med student?
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Many of life’s failures are people who did not realize how close they were to success when they gave up. |
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#9 | |
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Hall Of Fame
Join Date: Feb 2009
Posts: 1,670
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Quote:
My advice would be if your thinking about medicine . . . STOP !!! lol. |
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#10 | |
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Semi-Pro
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Quote:
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#11 | |
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Quote:
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#12 |
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New User
Join Date: Jun 2009
Posts: 12
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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?
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#13 |
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Join Date: Feb 2009
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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. |
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#14 | |
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Hall Of Fame
Join Date: Feb 2004
Posts: 2,256
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Quote:
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. |
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