# What is the volume occupied by 51.0 g of ammonia (NH_3) gas at STP?

May 24, 2016

The boiling point of ammonia is $- {33.34}^{\circ} \text{C}$. So, above the boiling point, we have STP, which is ${0}^{\circ} \text{C}$ and $\text{1 bar}$ (or $\text{1 atm}$, if your book is old).

Hence, ammonia is a gas at STP.

Assuming ideality, we use the ideal gas law as

$\setminus m a t h b f \left(P V = n R T\right)$

where:

• $P$ is the pressure. Let's use $\text{1 bar}$.
• $V$ is the volume in $\text{L}$.
• $n$ is the $\setminus m a t h b f \left(\text{mol}\right)$s of gas.
• $R = 0.083145 \left(\text{L"cdot"bar")/("mol"cdot"K}\right)$ is the universal gas constant for your units.
• $T$ is the temperature in $\text{K}$.

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

To get $n$, use the molar mass of ammonia to get:

51.0 cancel("g") xx ("1 mol NH"_3)/(17.0307 cancel("g NH"_3))

$=$ $\text{2.995 mols}$

Finally, solve for the volume.

color(blue)(V) = ((2.995 cancel("mols"))(0.083145 ("L"cdotcancel("bar"))/(cancel("mol")cdotcancel("K")))(273.15 cancel("K")))/cancel("1 bar")

$=$ $\textcolor{b l u e}{\text{68.01 L}}$

CHALLENGE: Can you use the density of ammonia gas, $0.761$ $g \text{/} L$, to solve for the volume using the mass of ammonia gas? You should get $67.02$ $L$. Does this mean ammonia is easier to compress than an ideal gas, or harder?