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### Topic: Balancing  (Read 25991 times)

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#### Shea

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##### Re: Balancing
« Reply #30 on: August 01, 2006, 05:18:07 PM »
Well, the other questions involved a product.  This one's different.

#### Borek

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##### Re: Balancing
« Reply #31 on: August 01, 2006, 05:25:11 PM »
Well, the other questions involved a product.  This one's different.

It doesn't matter. You always start with reaction equation and molar ratios it defines.
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#### Shea

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##### Re: Balancing
« Reply #32 on: August 01, 2006, 05:27:45 PM »
Well, how do I find the mass of F2 when the products don't matter and I only know that it's supposed to react with 100g of NaBr?

The program says the answer is 18.46g.  What steps did it take to get that answer?
« Last Edit: August 01, 2006, 05:36:58 PM by Shea »

#### Borek

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##### Re: Balancing
« Reply #33 on: August 01, 2006, 05:48:24 PM »
The program says the answer is 18.46g.

At least you know how to correctly use EBAS

100g NaBr - how many moles is it?

Look at the reaction equation. How many moles of F2 react with 1 mole of NaBr? How many moles F2 will react with known number of moles of NaBr?

It is identical with the calculations you did yesterday.
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#### Shea

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##### Re: Balancing
« Reply #34 on: August 01, 2006, 06:09:34 PM »
EBAS says 100g of NaBr is .97 moles.

That means the equation says 1 mole of F2 and .97 moles of NaBr will make 2NaF + Br2?

If its just 1 mole of F2, than thats just 38g?

#### Borek

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##### Re: Balancing
« Reply #35 on: August 01, 2006, 06:20:36 PM »
EBAS says 100g of NaBr is .97 moles.

I hope you know how to calculate it by yourself.

Quote
That means the equation says 1 mole of F2 and .97 moles of NaBr will make 2NaF + Br2?

No. Look at the reaction equation and coefficients there:

F2 + 2NaBr -> 2NaF + Br2

It reads: 1 mole of F2 reacts with 2 moles of NaBr yielding 2 moles of NaF and 1 mole of Br2. Now the question is: if 2 moles of NaBr react with 1 mole of F2, how many moles of F2 will react with 0.97 mole of NaBr?

It is described in the lecture I have pointed you to earlier tonight.
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#### Shea

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##### Re: Balancing
« Reply #36 on: August 01, 2006, 06:24:17 PM »
Yeah, I can calculate that myself.  I can do all the calculations, the hard part is just knowing what to calculate.

But, thanks to your rephrasing of the question, I'm not as confused anymore.

That's why EBAS says .4859 moles for F2.

Thats just a simple ratio.

#### Borek

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##### Re: Balancing
« Reply #37 on: August 01, 2006, 06:34:46 PM »
Thats just a simple ratio.

Told you so Once you understand that almost all stoichiometric questions are identical.
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