Question

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

Answer 1

The boiling point of ammonia is `-33.34^@ "C"`. So, above the boiling point, we have STP, which is `0^@ "C"` and `"1 bar"` (or `"1 atm"`, if your book is old).

Hence, ammonia is a gas at STP.

Assuming ideality, we use the ideal gas law as

`\mathbf(PV = nRT)`

where:

• `P` is the pressure. Let's use `"1 bar"`.
• `V` is the volume in `"L"`.
• `n` is the `\mathbf("mol")`s of gas.
• `R = 0.083145 ("L"cdot"bar")/("mol"cdot"K")` is the universal gas constant for your units.
• `T` is the temperature in `"K"`.

`V = (nRT)/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))`

`=` `"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")`

`=` `color(blue)("68.01 L")`

CHALLENGE: Can you use the density of ammonia gas, `0.761` `g"/"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?