Chemistry limiting reactants

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A student prepares phosphorous acid by reacting solid phosphorous triodide with water

PI3+H2O-->H3PO3+HI

The student needs to obtain 0.250L of phosphorous acid. (d=1.651g/mL). The procedure calls for a 45.0% excess of water and a yield of 75.0%. How much phosphorous triiodide should be weighed out?

My brain is fried from math today can someone help with the steps to solve this???

  • Chemistry limiting reactants -

    Balance the equation.
    PI3 + 3H2O ==> H3PO3 + 3HI

    I believe the problem assumes the H3PO3 prepared in this manner will be 100% purity; therefore, how many moles is in the product?
    1.651 g/mL x 250 mL = 412.75 grams.
    412.75/molar mass H3PO3 = 5.034 moles H3PO3.
    The amount of PI3 must also be 5.034 moles (since 1 mole PI3 produces 1 mole H3PO3). Convert that to grams PI3. Since it is only 75% efficient, divide the final # grams by 0.75.

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