Question #41855

The interesting thing about diphosphorus pentoxide is that its empirical formula is #"P"_2"O"_5#. Its molecular formula is actually #"P"_4"O"_10#.

This means that the answer will depend on which form is used in the chemical equation provided to you.

If your teacher or instructor used #"P"_2"O"_5# as the molecular formula, then you can say that

#"P"_2"O"_text(5(s]) + 3"H"_2"O"_text((l]) -> color(red)(2)"H"_3"PO"_text(4(aq])#

Notice that you have a #1:color(red)(2)# mole ratio between diphosphorus pentoxide and phosphoric acid. This tells you that for every mole of the former that reacts, the reaction will produce two moles of the latter.

You know that water is in excess, so you can ssume that all the diphosphorus pentoxide will react.

Well, if one mole of #"P"_2"O"_5# produces two moles of #"H"_3"PO"_4#, you can say that

#0.440color(red)(cancel(color(black)("moles P"_2"O"_5))) * (color(red)(2)color(white)(x)"moles H"_3"PO"_4)/(1color(red)(cancel(color(black)("mole P"_2"O"_5)))) = color(green)("0.880 moles H"_3"PO"_4#

If your chemical equation used #"P"_4"O"_10#, then you can say that

#"P"_2"O"_text(5(s]) + 6"H"_2"O"_text((l]) -> color(red)(4)"H"_3"PO"_text(4(aq])#

This time, one mole of #"P"_4"O"_10# produces four moles of #"H"_3"PO"_4#. Therefore, you get

#0.440color(red)(cancel(color(black)("moles P"_2"O"_5))) * (color(red)(4)color(white)(x)"moles H"_3"PO"_4)/(1color(red)(cancel(color(black)("mole P"_2"O"_5)))) = color(green)("1.76 moles H"_3"PO"_4#