1.1 What is Chemistry?

12.3. Limiting Reactant and Percent Yield http://www.ck12.org

Here is an interactive simulation to help you visualize reactions involving limiting reactants: http://phet.colorado
.edu/en/simulation/reactants-products-and-leftovers.

Limiting Reactants and Reaction Tables

Filling out a reaction table will help to double-check that you have found the correct limiting reactant, and it will
give you information about theoretical yields and the amounts that will be left over for any excess reactants.

Example 12.9

3.40 g of hydrogen gas and 7.16 g of nitrogen gas react to form gaseous ammonia. Assume that the reaction runs
until one of the reactants is fully consumed.

Answer:

First, write the balanced equation for this reaction and set up a reaction table. Fill in the amounts given in the
problem, and calculate the molar masses of each reaction component. Recall that the "change in moles" line is based
on the coefficients from the balanced equation:

N 2 (g)+ 3 H 2 (g)→2NH 3 (g)

TABLE12.6:NH3 Table Set Up

N 2 3H 2 2NH 3

Molar Mass 28.02 g/mol 2.02 g/mol 17.04 g/mol

Initial Mass 7.16 g 3.40 g 0 g

Initial Moles 0 mol

Change in Moles -x -3x +2x

Final Moles

Final Mass

Next, use molar masses to convert the known masses to moles:

7 .16 g N 2 (

1 mol N 2

28 .02 g N 2

) = 0 .256 mol N 2

3 .40 g H 2 (

1 mol H 2

2 .02 g H 2

) = 1 .68 mol H 2

Now that we have the moles of each reactant, we can divide by the coefficients from the balanced equation to
determine which will run out first.

0 .256 mol N 2

1

= 0. 256

1 .68 mol H 2

3

= 0. 560

Because the number for N 2 is lower, N 2 is the limiting reactant. Since this reaction is run until a reactant runs out,
we now know that the final mass of N 2 will be 0 grams. Putting this new information into the table, we get the
following: