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 > 2 kf

1) Molar mass:

K =  39.0983 g/mol
F₂ =  37.99 g/mol

moles ratio:

2 K + F₂   =   2 KF

2 x 39.0983 g K 37.99 g F₂
23.5 g K ( mass of F₂)

mass of F₂ = 23.5 x 37.99 / 2 x 39.0983

mass of F₂ = 892.765 / 78.1966

mass of F₂ = 11.4169 g

Therefore:

37.99 g F₂ 22.4 L ( at STP)
11.4169 g F₂ ( volume F₂ )

Volume F₂ = 11.4169 x 22.4 / 37.99

Volume F₂ = 255.73856 / 37.99

Volume F₂ = 6.731 L

Ais the correct answer

Particle exchange energy though elastic collision , hope this .