We'll first write the chemical equation for this reaction:
#"P"_4"O"_10(s) + 6"H"_2"O"(l) rarr 4"H"_3"PO"_4(aq)#
Let's now find the relative number of moles of phosphoric acid (#"H"_3"PO"_4#) using the coefficients of the equation, and the fact that #10.0# #"mol H"_2"O"# react:
#10.0cancel("mol H"_2"O")((4color(white)(l)"mol H"_3"PO"_4)/(6cancel("mol H"_2"O"))) = color(red)(6.67# #color(red)("mol H"_3"PO"_4#
Now, using the molar mass of #"H"_3"PO"_4# (calculated to be #98.00# #"g/mol"#), let's find the theoretical mass of phosphoric acid produced:
#color(red)(6.67)color(white)(l)cancel(color(red)("mol H"_3"PO"_4))((98.00color(white)(l)"g H"_3"PO"_4)/(1cancel("mol H"_3"PO"_4))) = color(blue)(653# #color(blue)("mol H"_3"PO"_4#
Thus, if the reaction goes to completion, there will be a yield of #color(blue)(653# #sfcolor(blue)("grams of phosphoric acid"#.
