# Question 4ceaa

Oct 19, 2015

$\text{11.3 L}$ at STP

#### Explanation:

Your question cannot be answered using the information you provided, so I'll have to make some assumptions.

SInce no mention of pressure and temperature was made, I'l assume that you're at STP conditions, which imply a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Now, ammonium nitrate, ${\text{NH"_4"NO}}_{3}$, will undergo decomposition to form nitrogen gas, oxygen gas, and water, according to the following balanced chemical equation

$2 {\text{NH"_4"NO"_text(3(aq]) -> 2"N"_text(2(g]) uarr + "O"_text(2(g]) uarr + color(red)(4)"H"_2"O}}_{\textrm{\left(g\right]}}$

Notice that you have a $2 : \textcolor{red}{4}$ mole ratio between ammonium nitrate and water. This means that the reaction will produce twice as many moles of water as you have moles of ammonium nitrate that undergo decomposition.

Now, use ammonium nitrate's molar mass to find how many moles you'd get in that many grams

20.0color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.043color(red)(cancel(color(black)("g")))) = "0.24987 moles"

Next, use the forementioned mole ratio to find how many moles of water would be produced when this many moles of ammonium nitrate undergo decomposition

0.24987color(red)(cancel(color(black)("moles NH"_4"NO"_3))) * (color(red)(4)" moles H"_2"O")/(2color(red)(cancel(color(black)("moles NH"_4"NO"_3)))) = "0.49974 moles H"_2"O"

To get the volume occupied by this many moles of water at STP, use the ideal gas law equation

$P V = n R T$

$V = \frac{n R T}{P}$

Use the STP values to get

$V = \left(0.49974 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm}}}}\right)$

$V = \text{11.342 L}$

Rounded to three sig figs, the number of sig figs you gave for the mass of ammonium nitrate, the answer will be

$V = \textcolor{g r e e n}{\text{11.3 L}}$

SIDE NOTE The same approach can be used regardless of the conditions you were given for pressure and temperature.

Once you get the moles of water, you can use whichever values you have.

Also, it is worth noting that the volume occupied by one mole of any ideal gas at STP is equal to 22.7 L - this is known as the molar volume of a gas at STP.

This means that when you're at STP, all you really need in order to calculate the volume of a gas is the number of moles. In this case, you would get

0.49974color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = color(green)("11.3 L")#