Limiting Reagents
Have you ever noticed that hot dogs are sold in packages
of 10, while the buns come in packages of eight? In this scenario,
the buns are the limiting reactant in the sense that they limit
the hot dog preparation to eight. The limiting reactant or reagent
is the one that is consumed first in the chemical reaction, and
its consumption halts the progress of the forward reaction.
When answering questions about limiting reagents on the
exam, your first step should always be to convert all the masses
you were given into moles. You should set up your table as you did
before, only now you’ll have two amounts and thus
two numbers of moles to get you started.
Let’s look at a specific question, involving the Haber
process. Basically, this is the process of making ammonia from the
reaction of nitrogen and hydrogen gases. The reaction is shown below:
| Molar mass |
(28.02) |
(2.02) |
(17.04) |
| Balanced equation |
N2 + |
3H2  |
2NH3 |
| No. of moles |
|
|
|
| Amount |
|
|
|
Suppose you have a total of 25.0 kg of nitrogen to react
with a total of 5.00 kg of hydrogen. What mass of ammonia can be
produced? Which reactant is the limiting reactant? What is the mass
of the reactant that’s in excess? Insert the masses in the correct
rows and find the number of moles of both.
| Molar mass |
(28.02) |
(2.02) |
(17.04) |
| Balanced equation |
N2 + |
3H2  |
2NH3 |
| No. of moles |
892 mol |
2475 mol |
|
| Amount |
25,000 g |
5000 g |
|
Start with nitrogen. You have 892 moles of it available,
and in order for the nitrogen to react completely with hydrogen,
you’d need 3(892 mol) = 2676 moles of hydrogen, which you don’t
have. Therefore, hydrogen is the limiting reagent. Now let’s answer
the other parts of the question. The mass of ammonia that can be
produced is limited by the amount of hydrogen, so do your calculations
based on the number of moles of hydrogen available. Your chart should
look like the one below:
| Molar mass |
(14.02) |
(2.02) |
(17.04) |
| Balanced equation |
N2 + |
3H2  |
2NH3 |
| No. of moles |
825 mol used |
2475 mol used |
therefore 1650 mol produced |
| Amount |
892 mol  23,117 g used
25,000g  |
5000 g |
1650 mol (17.04)
= 28,116 g produced |
