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13 December, 08:28

How many liters of fluorine gas, at 298 K and 0.98 atm, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem. 2K + F2 2KF

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  1. 13 December, 08:51
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    For the reaction 2 K + F2 - - > 2 KF,

    consider K atomic wt. = 39

    23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

    Now, following the ideal gas equation, PV = nRT

    P = 0.98 atm

    V = unknown

    n = 0.3015 moles

    R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)

    T = 298 K

    Solving for V,

    V = (nRT) / P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K) / (0.98 atm)

    solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.
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