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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
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number of mole of potassium=23.5/39.1 0.60 moles of potassium ratio of potassium to fluorine is 2:1 so 0.60 is halved so we get 0.30 moles of fluorine under standard condition fluorine occupies a volume of 24dm^3 0.30*24=7.2dm^3
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