For all you Chemistry buffs

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Tillek
02-02-2003, 06:46 PM
Problem:

How many grams of iodine trifluoride can be made from 5.00 grams of fluorine if there is an excess (more than enough) of iodine available?


thats exactly how the problem appear in the book. Does anybody know how to solve this? help please.

sled
02-02-2003, 06:51 PM
gimme the formula and ill do it

(eg: NaCl + Br --> NaBr + Cl (or whatever))

mongors
02-02-2003, 06:52 PM
That means that flourine is the limiting reagent and you have to use that equation (which I forget) to make a balanced equation. From that you can find out how many moles of iodine triflouride there is and then multiply by the atomic mass number to get the mass of the product.

Tillek
02-02-2003, 06:57 PM
well Iodine trifluoride is IF3 (3 is subscript). Fluorine is F2 (2 is subscript). Iodine is I2 (2 is subscript)

logmans
02-02-2003, 07:01 PM
that just means you can make 1 molecule for every 3 molecules of fluorine. Convert the 5g f to moles, divide by 3, then calculate the weight of that # of iodine trifluoride molecules. piece o cake

Tillek
02-02-2003, 07:04 PM
thanks.

Mystro
02-02-2003, 07:07 PM
32.2g, if I did this correctly

2I + 3F2 --> 2IF3

mass F2 = 5
molar mass F2 = 18.998 g/mol
mol F2 = 0.263


nIF3 2
------ = -
0.263F2 3

0.263 * 2 / 3 = 0.175

0.175 * molar mass IF3 (18.998 * 3 + 126.9)
= 32.2g


I *think* that should be it... I would've given you the obligatory "do your own hw" answer but I felt curious as to whether I could still do this or not, heh

(IIRC, not a limiting reagent question per se because there is only one substance given with mass)

sled
02-02-2003, 07:16 PM
Well this is how i would set it up on paper...
I think it's right..I rounded alot.


I2 + 3F2 --> 2IF3
m = 5g mol ratio = 2/3
M = 18.998 (2/3) x 0.2631mol
n = ? n = 0.1754mol

n = m/M m = nM
n = 5.00g/18.998g/mol m = 0.1754mol x 183.899 g/mol
n = 0.2631 mol m = 32.26 g

SegaRob
02-02-2003, 07:19 PM
Originally posted by Mystro
32.2g, if I did this correctly

2I + 3F2 --> 2IF3

mass F2 = 5
molar mass F2 = 18.998 g/mol
mol F2 = 0.263


nIF3 2
------ = -
0.263F2 3

0.263 * 2 / 3 = 0.175

0.175 * molar mass IF3 (18.998 * 3 + 126.9)
= 32.2g


I *think* that should be it... I would've given you the obligatory "do your own hw" answer but I felt curious as to whether I could still do this or not, heh

(IIRC, not a limiting reagent question per se because there is only one substance given with mass)

mistake 1:

2I + 3F2 --> 2IF3 should be I2 + 3F2 -> 2IF3.

mistake 2: molar mass of F2 is 38 (19 * 2) not 19

the rest seems correct


p.s. the correct answer is 16.13g

Mystro
02-02-2003, 07:21 PM
Originally posted by SegaRob


mistake 1:

2I + 3F2 --> 2IF3 should be I2 + 3F2 -> 2IF3.

mistake 2: molar mass of F2 is 38 (19 * 2) not 19

the rest seems correct


doh, totally missed the fact that iodine is diatomic, and I always forget to multiply molar mass in diatomic molecules :(

SegaRob
02-02-2003, 07:23 PM
np - sled did the same thing. vag chem newbs ;)

ptavv
02-02-2003, 07:23 PM
This sounds suspiciously like high school Chem...

Mystro
02-02-2003, 07:24 PM
I actually just had an exam on this material in Chem... The exam was ****ing brutal though because the teacher decided to put material that was in the book but was not taught nor on the review sheet :|

Mystro
02-02-2003, 07:24 PM
Originally posted by Ptavv
This sounds suspiciously like high school Chem...

dunno what grade he's in, but we did this in grade 11U chemistry

ptavv
02-02-2003, 07:25 PM
Originally posted by Mystro
I actually just had an exam on this material in Chem... The exam was ****ing brutal though because the teacher decided to put material that was in the book but was not taught nor on the review sheet :| Good luck in college.

SegaRob
02-02-2003, 07:26 PM
yea, that's definitely first-year stuff, kid's stuff ;)

ptavv
02-02-2003, 07:27 PM
Originally posted by Mystro
(IIRC, not a limiting reagent question per se because there is only one substance given with mass) The only way it's not a limiting reagent question is that they're telling you what the limiting reagent is and not asking you... Pretty much any chemistry question which involves "an excess" of one product and a "limited" amount of the second product is a limiting reagent-related question.

Mystro
02-02-2003, 07:29 PM
Originally posted by Ptavv
The only way it's not a limiting reagent question is that they're telling you what the limiting reagent is and not asking you... Pretty much any chemistry question which involves "an excess" of one product and a "limited" amount of the second product is a limiting reagent-related question.

I actually thought of that after I typed it... What I meant by limiting reagent is the fact that you don't have to find it

:)

Tillek
02-02-2003, 07:45 PM
gah! now im even more confused. I got 8.07 when I did it the first time and 32.2g when I did it the second time. oh and Mystro I wanted to know how to do it not the answer. please someone explain it.

SegaRob
02-02-2003, 07:48 PM
mystro already explained it well; he just missed a few things which i already noted. what more do you need?