At #"1.20 atm"#, and a temperature of #"27"^@"C"#, #"75.0 L NH"_3"# gas is produced by the reaction between #"N"_2"# gas and #"H"_2"# gas. How many moles and what volume of #"N"_2"# and #"H"_2"# are required to produce #"75.0 L NH"_3"#?

Balanced Equation

#"N"_2("g") +3"H"_2("g")##rarr##"2NH"_3"(g)"#

Use the ideal gas law with the equation #PV=nRT#, where #P# is pressure, #V# is volume, #n# is moles, #R# is the universal gas constant, and #T# is temperature in Kelvins.

Known/Given:

All Gases:

#P="1.20 atm"#

#T=27^"o""C"+273.15="300 K"#

#R="0.08206 L atm K"^(-1) "mol"^(-1)"#

Ammonia #"NH"_3"#

#V="75.0 L"#

Unknowns:

#"NH"_3:# moles

#"N"_2"# and #"H"_2:# moles and volume

Determine the moles Ammonia #"NH"_3"#.

#PV=nRT#

#n=(PV)/(RT)=(1.20color(red)cancel(color(black)("atm"))xx75.0color(red)cancel(color(black)("L")))/(0.082060color(red)cancel(color(black)("L"))xx color(red)cancel(color(black)("atm"))xxcolor(red)cancel(color(black)("K"))^(-1)xx"mol"^(-1)xx300color(red)cancel(color(black)("K")))="3.656 mol NH"_3"#

Determine the moles and volume of #"N"_2"#.

Multiply the moles #"NH"_3"# times the molar ratio of #"N"_2# and #"NH"_3"# from the balanced equation with #"mol N"_2"# in the numerator.

#3.656color(red)cancel(color(black)("mol NH"_3))xx(1"mol N"_2)/(2color(red)cancel(color(black)("mol NH"_3)))="1.828 mol N"_2"#

Determine the volume of #"1.828 mol N"_2"# at #"1.20 atm"# and #"300 K"# using the ideal gas law.

Rearrange the ideal gas law equation to isolate #V#. Plug in the known values and solve.

#V=(nRT)/P#

#V=(1.828color(red)cancel(color(black)("mol N"_2))xx0.08206"L"xxcolor(red)cancel(color(black)("atm"))xxcolor(red)cancel(color(black)("K"))^(-1)xxcolor(red)cancel(color(black) ("mol"))^(-1)xx300color(red)cancel(color(black)("K")))/(1.20color(red)cancel(color(black)("atm")))="37.50 L N"_2"#

Determine the moles and volume of #"H"_2"# at #"1.20 atm"# and #"300 K"#.

Multiply the moles #"NH"_3"# times the molar ratio of #"H"_2"# and #"NH"_3"# from the balanced equation with #"mol H"_2"# in the numerator.

#3.656color(red)cancel(color(black)("mol NH"_3))xx(3"mol H"_2)/(2color(red)cancel(color(black)("mol NH"_3)))="5.484 mol H"_2"#.

Determine the volume of #5.484"mol H"_2# at #"1.20 atm"# and #300"K"# using the ideal gas law.

Rearrange the ideal gas law equation to isolate #V#. Plug in the known values and solve.

#V=(nRT)/P#

#V=(5.484color(red)cancel(color(black)("mol H"_2))xx0.08206"L"xxcolor(red)cancel(color(black)("atm"))xxcolor(red)cancel(color(black)("K"))^(-1)xx color(red)cancel(color(black)("mol"))^(1-)xx300color(red)cancel(color(black)("K")))/(1.20color(red)cancel(color(black)("atm")))="112.5 L H"_2"#

A very important thing to notice here is that you're dealing with gases that are at the same pressure and temperature. This means that their mole ratios become equivalent to their volume ratio.

#PV = nRT => V/n = underbrace((RT)/P)_(color(blue)("constant"))#

So, for any two gases that take part in this reaction, you have

#V_1/n_1 = (RT)/P = V_2/n_2 => n_1/n_2 = V_1/V_2#

This means that you can determine the volume of nitrogen and hydrogen by

#75.0cancel("L"NH_3) * ("3 L "H_2)/(2cancel("L"NH_3)) = color(green)("113 L")#

and

#75.0cancel("L"NH_3) * ("1 L "N_2)/(2cancel("L"NH_3)) = color(green)("37.5 L")#

So, once you use the ideal gas law equation to get the number of moles of ammonia, all you have to do is use the mole ratios.

#PV = nRT => n = (PV)/(RT)#

#n_(NH_3) = (1.20cancel("atm") * 75.0cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 27.0)cancel("K")) = "3.657 moles"# #NH_3#

Rounded to three sig figs, you have

#n_(NH_3) = color(green)("3.66 moles")# #NH_3#

Therefore, you get

#3.657cancel("moles"NH_3) * ("3 moles "H_2)/(2cancel("moles"NH_3)) = color(green)("5.49 moles")# #H_2#

and

#3.657cancel("moles"NH_3) * ("1 mole "N_2)/(2cancel("moles"NH_3)) = color(green)("1.83 moles")# #N_2#