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

##### 1 Answer

The boiling point of ammonia is

Hence, ammonia is a **gas** at STP.

Assuming ideality, we use the **ideal gas law** as

#\mathbf(PV = nRT)# where:

#P# is thepressure. Let's use#"1 bar"# .#V# is thevolumein#"L"# .#n# is the#\mathbf("mol")# sof gas.#R = 0.083145 ("L"cdot"bar")/("mol"cdot"K")# is theuniversal gas constantfor your units.#T# is thetemperaturein#"K"# .

#V = (nRT)/P#

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*, *, 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?*