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]How many liters of fluorine gas can react with 32.0 grams of sodium metal at standard temperature and pressure? Show all of the work used to find your answer. 2Na + F2 2NaF
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n= m/M = 1,4mol n(Na):n(F2) = 2:1 n(F2) = 2*1,4 = 2,8 mol n= V/Vm -> V= n * Vm = 2,8 * 22,4 = 62,72 dm3
sorry made misstake n(F2) = 1/2 n(Na) = 0,7 mol V= 15,68 dm3
Thank you so much! i really did not get it. but now i understand it.
not a problem, glad to help
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