How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem.

2 K + F2 yields 2 KF

Question

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem. 

2 K + F2 yields 2 KF

anonymous · Answer

Potassium (K) = 39.098 g/mol

23.5g / 39.098 g/mol = 0.601 mol
ratio: 2:1
0.601 mol / 2 = 0.3005 mol
V = nRT/P = 22.4 L     -   one mole of any gas occupies a volume of 22.4 L
n = 0.3005 mol
R = 0.0821
T = 273.15 K
P = 1 atm
(0.3005 * 0.0821 * 273.15) / 1 = 6.73
V = 6.73 L

anonymous · Answer

Can someone tell me if my answer is correct, please?

anonymous · Answer

@ganeshie8

anonymous · Answer

@Cuanchi

cuanchi · Answer

yes its correct!

anonymous · Answer

Hey, I'm back, I was away from my computer for a while. Thanks! :)