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What mass of F2 is needed to produce 116 g of PF3 if the reaction has a 77.4% yield?

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    P + 3F2 ==> 2PF3
    mols PF3 = 116/molar mass PF3 = about 1.3.
    Convert mols PF3 to mols F2.
    1.3 mols PF3 x (3 mols F2/2 mols PF3) = 1.3 x (3/2) = 1.95 mols F2.
    How many g is that? 1.95 x molar mass F2 = about 74 g F2 (You shoud redo all of this more accurately.)
    Since this is only 77.4% yield, then
    74/0.774 = g F2 needed to produce 116 g F2.

    You could go another route. You could have said 77.4% of what number is 116 which gives 116/0.774 = about 149.9 and use that number for grams PF3, then calculate grams F2 needed to produce that. You should get the same answer either way.

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