Posted by Gweedo8 on Sunday, May 31, 2009 at 6:55pm.
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  DrBob222, Sunday, May 31, 2009 at 7:37pm
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.