# Ammonia reacts with oxygen according to the equation given below. If a "72.0-mL" sample of "NH"_3 gas is allowed to react with excess oxygen at room temperature, 25^@"C", and "1 atm", calculate the number of molecules of water produced?

## $4 \text{NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O} \left(l\right)$

Jul 9, 2017

$2.7 \cdot {10}^{21}$

#### Explanation:

The first thing that you need to do here is to use the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

to find the number of moles of ammonia present in your sample.

Rearrange the ideal gas law equation to solve for $n$

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

Plug in your values to find--do not forget that you need to convert the volume of the sample to liters and the temperature to Kelvin!

$n = \left(1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 72.0 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K}}}}\right)$

$n = {\text{0.0029414 moles NH}}_{3}$

Now, you know by looking at the balanced chemical equation

$4 {\text{NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O}}_{\left(l\right)}$

that for every $4$ moles of ammonia that take part in the reaction you get $6$ moles of water.

This means that your reaction will produce--oxygen gas is said to be in excess, so you don't have to worry about it being a limiting reagent.

0.0029414 color(red)(cancel(color(black)("moles NH"_3))) * overbrace(("6 moles H"_2"O")/(4color(red)(cancel(color(black)("moles NH"_3)))))^(color(blue)("given by the balanced chemical equation"))

$= \text{0.004412 moles H"_2"O}$

To find the number of molecules of water produced by the reaction, use the fact that it takes $6.022 \cdot {10}^{23}$ molecules of water in order to have exactly $1$ mole of water $\to$ this is given by Avogadro's constant.

You can thus say that your sample will contain

0.004412 color(red)(cancel(color(black)("moles H"_2"O"))) * overbrace((6.022 * 10^(23)color(white)(.)"molecules H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))^(color(blue)("Avogadro's number"))

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{2.7 \cdot {10}^{21} \textcolor{w h i t e}{.} \text{molecules H"_2"O}}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the pressure at which the reaction takes place.