Posted by Am I doing this Problem right? on Saturday, December 12, 2009 at 4:29pm.
Below is the question and right after is my work.
Excess fluorine (F2) reacts at 150 degrees C with bromine (Br2) to a compound BrFn (n = variable # i.e. n = 2 = F2). If 423 mL Br2 at 150 degrees C and 748 mmHg produced 4.20 g BrFn, what is n?
PV = nRT
n = [(.9842 atm)(.423L)]/[(.0821)(423.15)
n = .011983698 moles Br 2
Out of the 4.20 grams of BrFn 1.91499494 grams is Br2
I don't know what to do after here. What's my next step?
- Chemistry - bobpursley, Saturday, December 12, 2009 at 5:25pm
you know how many moles of Br2, so you know how many moles of Br (.02396)
Now you have 4.20gramsBr, of that, 4.20-1.915g is Fluorine., or .1204 moles
ratio of Fl/Br= .1204/.02396=5
- Chemistry - Am I doing this Problem right?, Saturday, December 12, 2009 at 5:31pm
Thank you so much for your help!
- Chemistry - DrBob222, Saturday, December 12, 2009 at 5:38pm
I would suggest you divide the 1.915 g Br2 by 2 to determine the number of grams of Br (not Br2), then subtract from 4.20 g BrFn to determine the grams F.
Then (g F/atomic mass F) = moles F
and (g Br/atomic mass Br) = moles Br.
Finally, determine the ratio of the moles of one to moles of the other. The easiest way to do that is to take the smaller number and divide it by itself (thereby making 1.00 for that element), then divide the other number by the same small number. You should end up with a set of small whole numbers, or at least one that can be rounded to a whole number.
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