Question

Percent yield problem:
A 20.0 L nickel container was charged with 0.841 atm of xenon gas and 1.33 atm of fluorine gas at 400 C. The xenon and fluorine react to form xenon tetrafluoride. What mass (g) of xenon tetrafluoride can be produced assuming 100 percent yield.

Answer 1

Let me get my periodic table, hold on.

Answer 2

I balanced teh equation and got Xe + 2F2 yields XeF4. Then I was trying to use the ideal gas law PV=nRT to find moles of F2 but I keep getting a number that seems too off.

Answer 3

Make sure all you numbers have the correct units. What value of R are you using?

Answer 4

I think you need to find the total moles of reactants first.

Answer 5

I get the feeling you need to solve a system of equations to get this.

Answer 6

I used 8.314 as my R

Answer 7

and on my hw it shows hints to tell u what to do step by step. So this was the 2nd step after balancing equation. However anytime i input my answer I got which is 148 it marks it wrong

Answer 8

Show me all the steps you did (or at least describe them), so I can see where you started and how you got to that number.

Answer 9

ok so I did PV = nRT and I substituted to get (8.314) ( 473 K) over (1.33 atm) ( 20 L) = 147.84

Answer 10

So that is an n=PV/RT calculation to find number of moles of fluorine, correct?

Answer 11

Might want to check your temperature: 400 ºC = 674 K.

Answer 12

Also, you'll want to change your units of pressure and volume, or use a different value for R.

Answer 13

R=8.3145 is in units of joule per mole per kelvin. If you want to use liters and atmospheres, use R = 0.0820575

Answer 14

ok thanks let me input those changes and try again and see if it works

Answer 15

so the atm and volume are still correct right

Answer 16

ok this makes soo much more sense i got 2.07 But since it says mole of F2 do i have to leave it as 2.07

Answer 17

I still think there is more too it than that. I got 0.481 using those numbers, but I'm a little rusty on chem, are you sure you can use the total volume of 20L even though the Fl is only part of that mixture?

Answer 18

oh i was thinking that but idk what volume to use so i just stuvk with that

Answer 19

wait omg ur rightttt!! its 0.481

Answer 20

I'm thinking partial pressures - partial volumes, but I'm not sure if that's valid. If it is, then you need to find total moles first using total pressure, then find the individual volumes of reactants to get the moles of each reactant.

Answer 21

the next step says to find teh moles of Xe

Answer 22

so we use teh same R and T and the pressure as 0.841 atm and temp as 674 K. right

Answer 23

Yes, you need to know the number of moles of both reactants to see how many moles of product is in the yield.

Answer 24

*flips through his old chem book*

Answer 25

ok so i must be doing smthg wrong. i keep getting big numbers. I got 3.28

Answer 26

Ah, this will come in handy.
"The mole fraction is equal to the ratio of partial pressures."

Answer 27

Ok, yes, I think the way to go is to use PV=nRT on the whole mixture to find total number of moles. The total pressure is the sum of the partial pressures.

Answer 28

Then, you can find the number of moles of each gas because they are in the same ratio as their partial pressures.

Answer 29

do u mind telling me what u get for moles of Xe

Answer 30

Working it now, just a sec.

Answer 31

Ok. Yes, that does work, moles of Fl = 0.481 and moles of Xe = 0.304

Answer 32

i reported you for not being in the chemistry section

Answer 33

iput the question in teh chem section but no one was asnwering any questions there anyways

Answer 34

so i put it here. U wud be wrong since stoicheomotery is math

Answer 35

well i used to do that and then i would get banned so its not fair..sorry i know..the mods are a holes and this would happen anyway

Answer 36

i asked an economics question which is math and i got banned

Answer 37

Heya - just a reminder to ask questions in the correct section :)

If you have trouble getting help (which does happen in the slower groups) please ask and provide a link in the chatbox.

Answer 38

ok sure next time

Answer 39

Cheers

Answer 40

Ok 0.304 is correct. The nxt hint was to find teh mole ration btwn Xe and F2 and it was 1 to . Now it says to find how many moles of Xe are required to compeltely react with F2?

Answer 41

Yeah, I was going to ask if you noticed that the fluorine was a limiting reactant.

Answer 42

Yeah so should I use a ratio of 0.305 mole Xe/ 0.481 mole F = (x moles Xe/ 1 mole F)

Answer 43

Not quite. The reactants are in a ratio of Xe : F :: 1 : 2, i.e. for every 2 moles of F you need 1 mole of Xe. If there are 0.481mol F, then that can only react with 0.2404mol Xe.

Answer 44

There is an excess of xenon that will not react (for practical purposes, this is a safety measure. Xenon is non-reactive under normal circumstances and is for the most part harmless. Fluorine gas is very dangerous, so you wouldn't want any excess of that floating around after you collected of the tetrafluoride that you wanted).

Answer 45

Sorry i copied and pasted the previus statement

Answer 46

Disregard that

Answer 47

Ok so now 0.61 moles F2 IS REQUIRED TO react with Xe.

Answer 48

The last hint is how many moles of XeF4 can be produced? Hpw do i determine this

Answer 49

The number of moles of the tetrafluoride equals the number of moles of xenon. They are in the same 1:1 ratio.

Answer 50

(Also note that 0.61 moles of F2 is not available)

Answer 51

oh ok. So the final asnwer i just multipled 0.2404 times its molecualr weigth and got 49.76 which is correct! :)

Answer 52

I got 49.836, but probably used more digits in the calculation.

Answer 53

THANK YOU SOOO MUCH!!! You explained everything sooo thoroughly!

Answer 54

And yeah I probably rounded my answer before urs. You were a great help. Im sorry that I gave u trouble since u had to look up chemistry notes

Answer 55

Rounds off to 49.8 g regardless....

You're welcome. That was a blast from the past there. I tutored some chem like that a year ago, but haven't done much extensively with it for many years.

Answer 56

Thank you so much for going through the trouble. I aprreciate it alot! :)

Answer 57

You must have been a great chem tutor

Answer 58

btw in teh begining Idk why I used the same values as u for PV=nRT and kept getting large # instead of decimals like u. Im not sure what I mustve done wrong

Answer 59

You did start with the wrong R for the units, also you had the temperature too low a couple times.

Answer 60

Or it might have just been a calculator key punch error. *shrugs*

Answer 61

ok thanks!!! I'll be careful next time

Answer 62

thanks again

Answer 63

u did use 20 as ur volume right?

Answer 64

or did u convert that to smthg?