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Manganese(III) fluoride, MnF3, can be prepared by the following reaction:
2MnI2(s) + 13F2(g) → 2MnF3(s) + 4IF5(l)

If the percentage yield of MnF3 is always approximately 51%, how many grams of MnF3 can be expected if 40.0 grams of each reactant is used in an experiment?

  • chemistry -

    You have to convert each 40 grams to moles of reactants.

    Then, using the mole ratio 2:13, determine which is the limiting reactant. I assume you can do that. Then, say the limiting reactant is F2 (It probably wont be the real limiting reactant here), and you had XXX moles of it to react. Then moles of MnF3 you get will be .50*2/13*mole of F2 you really had.

  • chemistry -

    Oh yes, then convert those moles of MnF3 to grams.

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