Question

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal?

2 K + F2 ---> 2 KF

Show the work or explain somehow if possible por favor(:

Answer 1

Potassium 23.5g/39.0983g/mol = 0.601mol
The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol

Therefore 0.3005mol of F2 is needed to find liters use formula
V = nRT/P (V)Volume = 22.41L
(T)Temperature = 273K or 0.0 Celsius
(P)Pressure = 1.0atm
(R)value is always .08206 with atm
n = 0.3005moles

(273)(.08206)(0.3005)/1 = V
V = 6.7319 Liters
confirm for me zbay

Answer 2

alright you should be able to do this with a stocomic conversion because it's at STP,

\[23.5 grams of K (\frac{1 mole K}{39.1 g K}) (\frac{1 mole F_2}{2 mole K}) =.301 mole F_2\]

Now that we figured out that we can plug .301 mole Florine into the Ideal Gas law equation and get our answer.

\[PV=nRT\]

we need to solve for V

\[V=\frac{nRT}{P}\]

replace the values

\[V=\frac{(.301 mole)(293.15 K).08206}{1 atm}\]

and we get

V=7.23L

Answer 3

We have a disagreement on standard temp it looks like we did the rest the same

Answer 4

According to IUPAC standard temperature is 273.15 based on google search and according to my first year college prof :P

Answer 5

Ok i was using SATP, my bad

Answer 6

No problem thanks for refreshing my mind on these problems

Answer 7

Yea that was a good one to keep us sharp on some basic chem skills that are a bit rusty. Thats why i enjoy this site.