Chemistry

Manganese trifluoride, MnF3,can be prepared by the following reaction.
2 MnI2 (s) + 13 F2 (g) ---> 2 MnF3(s) + 4 IF5(l)
What is the minimum number of grams of F2 that must be used to react with 12.0 g of MnI2 if the overall yield of MnF3 is no more than 75% of theory?

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  1. 12.0g MnI2/0.75 = ? g MnI2 to start accounting for the 75% yield = approx 16

    mols MnI2 = grams/molar mass = 16/about 300 = ?
    Convert to mols F2. That's
    ? mols MnI2 x (13mols F2/2 mols MnI2) = ? x 13/2 = x mols MnI2.
    Convert to grams. g = mols x molar mass

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