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

1 Answer
May 24, 2016

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?