# Balance the following reaction: NH4NO3 + Na3PO4 –> (NH4)3PO4 + NaNO3 If we started with 30.0 grams of ammonium nitrate and 50.0 grams of sodium phosphate, which reactant is limiting? What is the mass of each product that can be formed? What mass of t…

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 ##”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”##

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