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. 2K + F2 2KF

For the reaction 2 K + F2 - - > 2 KF,

consider K atomic wt. = 39

23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

Now, following the ideal gas equation, PV = nRT

P = 0.98 atm

V = unknown

n = 0.3015 moles

R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)

T = 298 K

Solving for V,

V = (nRT) / P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K) / (0.98 atm)

solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.