# Question c31e8

Jan 26, 2016

$\text{0.353 g}$

#### Explanation:

The first thing to do here is write the balanced chemical equation for this decomposition reaction. When heated, ammonium nitrate, ${\text{NH"_4"NO}}_{3}$, will decompose to give nitrous oxide, $\text{N"_2"O}$, and water vapor.

${\text{NH"_4"NO"_text(3(s]) stackrel(color(red)("heat"color(white)(aa)))(->) "N"_2"O"_text((g]) + 2"H"_2"O}}_{\textrm{\left(g\right]}}$

Now, the thing to remember about ideal gases under STP (Standard Temperature and Pressure) conditions, which are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, that one mole of a gas occupies exactly $\text{22.7 L}$ - this is known as the molar volume of a gas at STP.

So, if you're working under STP conditions, one mole of any ideal gas occupies a volume of $\text{22.7 L}$. This means that the given volume of nitrous oxide will contain

0.100 color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.7 color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas")) = "0.004405 moles"

Now look at the balanced chemical equation. Notice that you have a $1 : 1$ mole ratio between ammonium nitrate and nitrous oxide.

This means that the reaction will produce one mole of nitrous oxide for every one mole of ammonium nitrate that undergoes decomposition.

Since the reaction produced $0.004405$ moles of $\text{N"_2"O}$, it follows that

0.004405 color(red)(cancel(color(black)("moles N"_2"O"))) * ("1 mole NH"_4"NO"_3)/(1color(red)(cancel(color(black)("mole N"_2"O")))) = "0.004405 moles NH"_4"NO"_3

Now all you have to do is use ammonium nitrate's molar mass, which gives you the mass of one mole of ammonium nitrate, to find how many grams would contain this many moles.

0.004405 color(red)(cancel(color(black)("moles NH"_4"NO"_3))) * "80.043 g"/(1color(red)(cancel(color(black)("mole NH"_4"NO"_3)))) = color(green)("0.353 g")#

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the nitrous oxide.

SIDE NOTE Many textbooks and online sources still use STP conditions as being a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22. 4L}$.

If this is the value you're supposed to use for the molar volume of a gas at STP, simply redo the calculations using $\text{22.4 L}$ instead of $\text{22.7 L}$.