# What is the volume of nitrogen that 50.0 g of ammonium nitrate produces when it decomposes? What is the mass of water formed?

Dec 19, 2017

The reaction produces 22.5 g of water and 14.2 L of nitrogen at STP.

#### Explanation:

The equation for the reaction is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 80.04 \textcolor{w h i t e}{m m m m m m m m m m m m m m} 18.02$
$\textcolor{w h i t e}{m m} \text{2NH"_4"NO"_3"(s)" → "2N"_2"(g)" + "O"_2"(g)" + "4H"_2"O(g)}$

A. Volume of nitrogen

${\text{ Moles of N"_2 = 50.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 color(red)(cancel(color(black)("mol NH"_4"NO"_3))))/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3))))× ("2 mol N"_2)/(2 color(red)(cancel(color(black)("mol NH"_4"NO"_3)))) = "0.6247 mol N}}_{2}$

You don't give the pressure and temperature at which the nitrogen is formed, so I shall assume STP (1 bar and 0 °C).

The molar volume of an ideal gas at STP is 22.71 L.

${\text{Volume of N"_2 = 0.6247 "mol N"_2 × "22.71 L N"_2/(1 "mol N"_2) = "14.2 L N}}_{2}$

B. Mass of water

$\text{ Moles of N"_2 = 50.0 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 color(red)(cancel(color(black)("mol NH"_4"NO"_3))))/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3))))× ("4 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol NH"_4"NO"_3)))) = "1.249 mol H"_2"O}$

$\text{Mass of H"_2"O" = 1.249 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) ="22.5 g H"_2"O}$