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4 December, 18:39

What mass of f2 is needed to produce 120. g of pf3 if the reaction has a 78.1% yield?

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  1. 4 December, 20:34
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    write the equation for the reaction

    that is 6 F2 + P4 = 4 PF3

    find the theoretical mass that is

    let the theoretical yield be represented by y

    theoretical yield = 78.1/100 = 120/y

    y = 153.6 grams

    find the number of moles of PF3

    moles = mass/molar mass

    = 153.6/87.97 = 1.746 moles

    by use of mole ratio between F2 : PF3 which is 6:4 the moles of F2 is therefore = 1.746 x 6/4 = 2.62 moles

    mass = moles x molar mass

    = 1.746 moles x38 g/mol = 99.6 grams
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