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.
2K + F2 yields 2KF

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

Answer 2

Hope it cleared your doubt