# Question 69942

Feb 19, 2017

$\text{1.21 g}$

#### Explanation:

The first order of business here is to figure out the number of moles of dinitrogen oxide produced by the reaction.

To calculate the number of moles of dinitrogen gas present in that $\text{0.240 L}$ sample, use the fact that under STP conditions, which are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, $1$ mole of any ideal gas occupies $\text{22.7 L}$.

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SIDE NOTE Keep in mind that most sources and textbooks still use a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$ as the definition of STP conditions.

In this case, one mole of any ideal gas occupies $\text{22.4 L}$. If this is the value for the moalr volume of a gas at STP given to you, simply redo the calculations using $\text{22.4 L}$ instead of $\text{22.7 L}$.

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In your case, the sample of dinitrogen oxide gas will contain

0.240 color(red)(cancel(color(black)("L"))) * ("1 mole N"_2"O")/(22.7color(red)(cancel(color(black)("L")))) = "0.01057 moles N"_2"O"

Now take a look at the balanced chemical equation that describes this decomposition reaction

${\text{NH"_ 4"NO"_ (3(s)) -> "N"_ 2"O"_ ((g)) + 2"H"_ 2"O}}_{\left(g\right)}$

Notice that every mole of ammonium nitrate that undergoes decomposition produces $1$ mole of dinitrogen oxide. This means that the reaction consumed

0.01057 color(red)(cancel(color(black)("moles N"_2"O"))) * ("1 mole NH"_4"NO"_3)/(1color(red)(cancel(color(black)("mole N"_2"O")))) = "0.01057 moles NH"_4"NO"_3#

To convert this to grams, use the molar mass of ammonium nitrate

$0.01057 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_4"NO"_3))) * "80.043 g"/(1color(red)(cancel(color(black)("mole NH"_4"NO"_3)))) = color(darkgreen)(ul(color(black)("1.21 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the volume of dinitrogen oxide produced by the reaction.