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.

2K + F2 yields 2KF

Hope it cleared your doubt

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