A #30*L# volume of dinitrogen reacts quantitatively with a #60*L# dihydrogen to give ammonia. What volume of ammonia results under the same conditions, and what is the volume of the excess reagent?

1 Answer
Feb 3, 2017

Answer:

By applying the stoichiometry of the equation.

Explanation:

Under conditions of equivalent temperature and pressure, the volume of a gas is proportional to its molar quantity. Here we have 30L of dinitrogen, and 60L of dihydrogen. Clearly, dinitrogen gas is the reagent in excess, and only 20L of the dinitrogen can react with the dihydrogen, proportional to the molar quantity of #H_2#.

Continuing the proportionality, if all the dinitrogen reacted, then 40L of ammonia would result.

We can go farther back and make the assumption of ideality:

#n=(PV)/(RT)#, and thus #npropV# if #T# and #P# are constant, which was explicitly stated in the problem. Do you agree?

These quantitative yields are impossibly high, and do not reflect the reality of the industrial process. Are you happy with this explanation?