Question #faabd

Start with the balanced chemical equation for the combustion of ammonia

#color(red)(4)NH_(3(g)) + color(green)(3)O_(2(g)) -> 2N_(2(g)) + 6H_2O_((g))#

Notice that you have a #color(red)(4):color(green)(3)# mole ratio between ammonia and oxygen. This means that, regardless of how many moles of ammonia react, you need 3/4 fewer moles of oxygen for the reaction to take place.

Because the pressure and temperature of the reaction are kept constant, you have, using the ideal gas law equation

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

This means that the mole ratios that exist between the compounds that take part in the reaction are equivalent to volume ratios.

In this case, regardless of how many cubic centimeters of ammonia you have, you'll always need 3/4 fewer cubic centimeters of oxygen for the reaction to take place.

Since you have #"100 cm"^3# of ammonia, you only need

#100cancel("cm"^3NH_3) * (color(green)(3)"cm"^3O_2)/(color(red)(4)cancel("cm"^3NH_3)) = "75 cm"^3# #O_2#

This tells you that ammonia acts as the limiting reagent, i.e. it doesn't allow for all the oxygen to react. As a result, you'll be left with excess oxygen.

In order to determine the volumes of nitrogen gas and of water vapor produced, you use the volume ratios that exist between ammonia and the products.

#100cancel("cm"^3NH_3) * ("2 cm"^3N_2)/(4cancel("cm"^3NH_3)) = "50 cm"^3# #N_2#

and

#100cancel("cm"^3NH_3) * ("6 cm"^3H_2O)/(4cancel("cm"^3NH_3)) = "150 cm"^3# #H_2O#

Therefore, after the reaction is complete, you'll be left with #"25 cm"^3# of oxygen, #"50 cm"^3# of nitrogen, and #"150 cm"^3# of water vapor.