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.

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You use the atomic number to find out how many protons and neutrons there are in the atom. you use the atomic mass to find out how many protons and electrons there are in the atom.