The volume of fluorine gas required to react with 2.67 g of calcium bromide to f
ID: 617535 • Letter: T
Question
The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41Explanation / Answer
Balanced equation: F2(g) + CaBr2(s) ---> CaF2 + Br2 ===> 1 mol F2 = 1 mol CaBr2 Calculation: 2.67 g CaBr2 x (1 mol CaBr2 / 199.886 g CaBr2) x (1 mol F2 / 1 mol CaBr2) = 0.0133576 mol F2 Assuming F2 is an ideal gas at these conditions: PV = nRT Solving for V: V = (n x R x T) / P where, n = 0.0133576 mol R = 0.08206 atm·L/mol·K T = 41 C + 273.15 = 314.15 K P = 4.31 atm Substituting numbers into V = (n x R x T) / P: V = (0.0133576 mol x 0.08206 atm·L/mol·K x 314.15 K) / 4.31 atm = 0.079895 L = 79.895 mL
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