For the combustion reaction #4"NH"_3(g) + 5"O"_2(g) -> 4"NO"(g) + 6"H"_2"O"(g)#, how many liters of #"O"_2(g)# would react with #"500 L"# of #"NH"_3# at STP?

#"625 L O"_2#

This is a limiting reactant problem using gases (assumed ideal).

Given #"500 L"# of #"NH"_3#, since we are at STP (presumably, #"1 atm"# and #0^@ "C"#), we can determine the molar volume of #"NH"_3# by assuming it is an ideal gas.

#PV = nRT#

#V/n = (RT)/P#

#= (("0.082057 L"cdotcancel"atm""/mol"cdotcancel"K")(273.15 cancel"K"))/(cancel"1 atm")#

#=# #"22.41 L/mol"#

Therefore, we have

#500 cancel("L NH"_3) xx "1 mol"/(22.41 cancel("L NH"_3(g)))#

#=# #"22.31 mols NH"_3#

From the reaction given, there needs to be #"5 mols O"_2(g)# for every #"4 mols NH"_3(g)#. Therefore:

#22.31 cancel("mols NH"_3) xx ("5 mols O"_2)/(4 cancel("mols NH"_3))#

#=# #"27.88 mols O"_2#

Since we are also assuming #"O"_2(g)# is an ideal gas, we utilize the same molar volume to get:

#27.88 cancel("mols O"_2(g)) xx "22.41 L"/cancel("mol O"_2(g))#

#=# #color(blue)("625 L O"_2(g))#

You could also have gone straight from the volume and used the mol ratio by ASSUMING that both #"O"_2(g)# and #"NH"_3(g)# are ideal gases:

#"500 L"# #cancel("NH"_3) xx (5 cancel"mols" "O"_2)/(4 cancel"mols" cancel("NH"_3))#

#= color(blue)("625 L O"_2(g))#