How many moles of Xe required to react completely with F2?
A 20L nickel was charged with .813 atm of xenon gas and 1.29atm of fluorine gas at 400 degree C. The xenon and fluorine react to form Xenon tetrafluoride. What mass of Xenon terafluoride can be produced assuming 100% yield?
Q1: ? grams of XeF4
Q2: how many moles of Xe required to react completely with F2?
how many moles of Xe required to react completely with F2?
2011-12-23 4:36 pm
回答 (1)
2011-12-23 11:14 pm
✔ 最佳答案
Xe + 2 F2 → XeF4PV = nRT
n = PV / RT = (0.813 atm) x (20 L) / ((0.08205746 L atm /K mol) x (400 + 273K)) = 0.294 mol Xe
n = (1.29 atm) x (20 L) / ((0.08205746 L atm /K mol) x (400 + 273K)) = 0.467 mol F2
0.467 mole of F2 would react completely with 0.467 x (1/2) = 0.234 mole of Xe, but there is more Xe present than that, so Xe is in excess and F2 is the limiting reactant.
(0.467 mol F2) x (1/2) x (207.2872 g XeF4/mol) = 48 g XeF4
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