Question

How many liters of fluorine gas, at 298 K and 0.98 atm, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem.
2 K + F 2 KF (4 points)

Answer 1

This is a simple stoichiometry problem. Use dimensional analysis to gonvert to moles of desired.

Here's the plan:
given --> want
hence,

grams of K --> moles of K --> moles of F

Now, using all the information clearly stated, use the ideal gas law formula (PV = nRT) to solve for volume [HINT: use algebra to rearrenge the equation and solve for V]


p = pressure
v = volume
n = moles
R = gas constant (you can find this value in your textbook)
T = temperature (usually in units of K)

Answer 2

pv=nrt

Answer 3

moles(n) of potassium first :23.5/molar mass of K(39)=602.564E-3
moles of fluorine second :mole of K/2, in this case=301.28205E-3
p=0.98atm
R=gas constant,lookout for the units(0.0821( L atm/k/mol)
Temperature=298K
v=?
plug in the values and have fun
Answer=(298*301.28205E-3*0.0821)/0.98=7.52153 L

Answer 4

@telijahmed you should of given him step=by=step instructions rather than given him the answer.

Answer 5

Saved my life with this answer :)