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 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 )
