# Question 3e282

Mar 11, 2014

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 ${\text{NH}}_{4}$, ${\text{NO}}_{3}$, and ${\text{PO}}_{4}$ as single units.

Start with the most complicated formula — ${\text{(NH"_4")"_3"PO}}_{4}$. Put a 1 in front of it.

${\text{NH"_4"NO"_3 + "Na"_3"PO"_4 → color(red)(1)("NH"_4")"_3"PO"_4 + "NaNO}}_{3}$

Then balance ${\text{NH}}_{4}$. Put a 3 in front of ${\text{NH"_4"NO}}_{3}$.

$\textcolor{t e a l}{3} {\text{NH"_4"NO"_3 + "Na"_3"PO"_4 → color(red)(1)("NH"_4")"_3"PO"_4 + "NaNO}}_{3}$

Balance ${\text{PO}}_{4}$. Put a 1 in front of ${\text{Na"_3"PO}}_{4}$.

$\textcolor{t e a l}{3} {\text{NH"_4"NO"_3 + color(blue)(1)"Na"_3"PO"_4 → color(red)(1)("NH"_4)_3"PO"_4 + "NaNO}}_{3}$

Balance $\text{Na}$. Put a 3 in front of ${\text{NaNO}}_{3}$.

$\textcolor{t e a l}{3} {\text{NH"_4"NO"_3 + color(blue)(1)"Na"_3"PO"_4 → color(red)(1)("NH"_4)_3"PO"_4 + color(orange)(3)"NaNO}}_{3}$

The balanced equation is

$\textcolor{red}{\text{3NH"_4"NO"_3 + "Na"_3"PO"_4 → "(NH"_4")"_3"PO"_4 + "3NaN} {O}_{3}}$

IDENTIFY THE LIMITING REACTANT

The molar masses are

$\text{NH"_4"NO"_3= "80.04 g/mol}$; $\text{Na"_3"PO"_4 = "163.94 g/mol}$

$\text{(NH"_4")"_3"PO"_4 = "149.09 g/mol}$; $\text{NaNO"_3 = "84.99 g/mo} l$

Mass of ${\text{(NH"_4")"_3"PO}}_{4}$ from ${\text{NH"_4"NO}}_{3}$:

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 ${\text{Na"_3"PO}}_{4}$:

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

${\text{NH"_4"NO}}_{3}$ 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 ${\text{NaNO}}_{3}$:

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 ${\text{NH"_4"NO}}_{3}$ remaining is zero, since the reaction uses up all the limiting reactant.

MASS OF EXCESS REACTANT REMAINING

Mass of ${\text{Na"_3"PO}}_{4}$ 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 $\text{Na"_3"PO"_4 " remaining = 50.0 g - 20.5 g = 29.5 g}$