This is a long question. It has a long answer.
I must point out that all the substances in the equation are soluble in water, so there is actually No Reaction.
If there were a reaction, this is how you would attack the problem.
BALANCE THE EQUATION:
Count the polyatomic ions #"NH"_4#, #"NO"_3#, and #"PO"_4# as single units.
Start with the most complicated formula — #"(NH"_4")"_3"PO"_4#. Put a 1 in front of it.
#"NH"_4"NO"_3 + "Na"_3"PO"_4 → color(red)(1)("NH"_4")"_3"PO"_4 + "NaNO"_3#
Then balance #"NH"_4#. Put a 3 in front of #"NH"_4"NO"_3#.
# color(teal)(3)"NH"_4"NO"_3 + "Na"_3"PO"_4 → color(red)(1)("NH"_4")"_3"PO"_4 + "NaNO"_3#
Balance #"PO"_4#. Put a 1 in front of #"Na"_3"PO"_4#.
# color(teal)(3)"NH"_4"NO"_3 + color(blue)(1)"Na"_3"PO"_4 → color(red)(1)("NH"_4)_3"PO"_4 + "NaNO"_3#
Balance #"Na"#. Put a 3 in front of #"NaNO"_3#.
# color(teal)(3)"NH"_4"NO"_3 + color(blue)(1)"Na"_3"PO"_4 → color(red)(1)("NH"_4)_3"PO"_4 + color(orange)(3)"NaNO"_3#
The balanced equation is
#color(red)("3NH"_4"NO"_3 + "Na"_3"PO"_4 → "(NH"_4")"_3"PO"_4 + "3NaN"O_3)#
IDENTIFY THE LIMITING REACTANT
The molar masses are
#"NH"_4"NO"_3= "80.04 g/mol"#; #"Na"_3"PO"_4 = "163.94 g/mol"#
#"(NH"_4")"_3"PO"_4 = "149.09 g/mol"#; #"NaNO"_3 = "84.99 g/mo"l#
Mass of #"(NH"_4")"_3"PO"_4# from #"NH"_4"NO"_3#:
#30.0 cancel("g NH₄NO₃") × (1 cancel("mol NH₄NO₃"))/(80.04 cancel("g NH₄NO₃")) × (1 cancel("mol (NH₄)₃PO₄"))/(3 cancel("mol NH₄NO₃")) ×#
#("149.09 g (NH"_4")"_3"PO"_4)/(1 cancel("mol (NH₄)₃PO₄")) = "18.6 g (NH"_4")"_3"PO"_4#
Mass of #("NH"_4")"_3"PO"_4# from #"Na"_3"PO"_4#:
#50.0 cancel("g Na₃PO₄")× (1 cancel("mol Na₃PO₄"))/(163.94 cancel("g Na₃PO₄")) × (1 cancel("mol (NH₄)₃PO₄"))/(1 cancel("mol Na₃PO₄")) ×#
#("149.09 g (NH"_4")"_3"PO"_4)/(1 cancel("mol (NH₄)₃PO₄")) = "45.5 g (NH"_4")"_3"PO"_4#
#"NH"_4"NO"_3# is the limiting reactant, because it gives the smaller amount of product.
MASS OF EACH PRODUCT
Mass of #("NH"_4")"_3"PO"_4 = "18.6 g"#
Mass of #"NaNO"_3#:
#30.0 cancel("g NH₄NO₃") × (1 cancel("mol NH₄NO₃"))/(80.04 cancel("g NH₄NO₃")) × (3 cancel("mol NaNO₃"))/(3 cancel("mol NH₄NO₃")) ×#
#("84.99 g NaNO"_3)/(1 cancel("mol NaNO₃")) = "31.9 g NaNO"_3#
MASS OF LIMITING REACTANT REMAINING
The mass of #"NH"_4"NO"_3# remaining is zero, since the reaction uses up all the limiting reactant.
MASS OF EXCESS REACTANT REMAINING
Mass of #"Na"_3"PO"_4# reacted:
#30 cancel("g NH₄NO₃") × (1 cancel("mol NH₄NO₃"))/(80.04 cancel("g NH₄NO₃")) × (1 cancel("mol Na₃PO₄"))/(3 cancel("mol NH₄NO₃")) ×#
#("163.94 g Na"_3"PO"_4)/(1 cancel("mol Na₃PO₄")) = "20.5 g (NH"_4")"_3"PO"_4#
The mass of #"Na"_3"PO"_4 " remaining = 50.0 g - 20.5 g = 29.5 g"#
