# Question d83e7

Jul 19, 2017

Here's what I got.

#### Explanation:

Since you can't really have $3.01$ molecules of nitrogen gas, I'll assume that you're actually dealing with $3.01 \cdot {10}^{23}$ molecules of nitrogen gas, the equivalent of

3.01 * 10^(23)color(red)(cancel(color(black)("molecules N"_2))) * overbrace("1 mole N"_2/(6.02 * 10^(23)color(red)(cancel(color(black)("molecules N"_2)))))^(color(blue)("Avogadro's constant"))

$= {\text{0.500 moles N}}_{2}$

Now, nitrogen gas and hydrogen gas react to form ammonia according to the following balanced chemical equation

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH}}_{3 \left(g\right)}$

Notice that it takes $1$ mole of nitrogen gas to produce $2$ moles of ammonia, which means that your reaction will produce--keep in mind that the nitrogen gas reacts completely!

0.500 color(red)(cancel(color(black)("moles N"_2))) * "2 moles NH"_3/(1color(red)(cancel(color(black)("mole N"_2)))) = "1.00 moles NH"_3#

Under Standard Conditions for Pressure and Temperature, $1$ mole of any ideal gas occupies $\text{22.7 L}$ $\to$ this is known as the molar volume of a gas at STP.

You can thus say that your reaction will produce

$1.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_3))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole NH"_3)))))^(color(blue)("under STP conditions")) = color(darkgreen)(ul(color(black)("22.7 L}}}}$

The answer is rounded to three sig figs, the number of significant figures you have for the number of molecules of nitrogen gas.