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If 150.0 grams of sodium phosphate react with 200.0 grams of iron (II) sulfate, how many grams of iron (II) ph

by Guest32836  |  earlier

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If 150.0 grams of sodium phosphate react with 200.0 grams of iron (II) sulfate, how many grams of iron (II) phosphate will be formed and what is the limiting reactant?

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  1. 3 FeSO4  & 2 Na3PO4 --> 1 Fe3(PO4)2  & 3 Na2SO4

    3 FeSO4 @ 151.908 g/mol  = 455.7 grams FeSO4 react

    2 Na3PO4 @ 163.94 g/mol = 327.9 grams Na3PO4 react

    1 Fe3(PO4)2 = 357.5g/mole is produced

    --------------------------------------...

    @ 150g /200 g they used 0.75 as many grams of Na3PO4 as FeSO4

    @ 327.9g / 455.7 they only needed 0.72 as many grams of Na3PO4 as FeSO4

    so to answer your second question, they  added too much Na3PO4

    Your limiting reagent is: FeSO4

    ====================================

    how many grams of iron (II) phosphate will be formed....

    ? g Fe3(PO4)2 from:

    200.0g FeSO4 @ 357.5g Fe3(PO4)2 / 455.7 g FeSO4 =

    your first answer: 156.9 grams of Fe3(PO4)2 will be produced

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