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Many of the General Chemistry students I have encountered over my thirty years of teaching have had trouble determining the limiting reactant in stoichiometric problems. The student can perform the correct molar conversions but then often will get confused about what quantities to compare in order to determine the limiting reactant. The following step-by-step technique combines previous problem-solving methods outlined in "Stoichiometry Set-Up Method" with additional steps, and aims to provide the student with appropriate guidance throughout a limiting reactant calculation.

Steps for determining the limiting reactant

  1. Write a balanced chemical equation.
  2. Write what is given and what needs to be calculated on the equation as follows:
    1. Data in moles go under the equation.
    2. Data in other units go above the equation.
  3. Draw a box containing two lines underneath any one of the reactants. This will serve as the chosen reactant for future calculations.
    1. Label the top line “Have.”
    2. Label the bottom line “Needed” and put an “X” on the line.
  4. Draw a path, using arrows, from the given data to the box under your chemical equation.
    1. Separate vertical and horizontal arrows.
    2. Horizontal arrows must go under the equation.
  5. Using the given quantity of the chosen reactant, calculate its number of moles and write the answer on the line labeled “Have.”
  6. Using the given quantity from the other reactant, calculate the number of moles needed for the chosen reactant and write the answer on the line labeled “Needed.”
  7. Determining the limiting reactant: If there is more reactant on the “Have” line than is identified on the “Needed” line, that reactant is in excess. If there is less of the chosen reactant than the quantity needed to completely react, then the chosen reactant is the limiting reactant.

Putting the steps into practice

The following problem illustrates how to use the technique described above with your students. When I teach limiting reactant calculations using this method, I write each of the following steps on the board as I walk the students through the following problem. I encourage you to do the same, or provide your students with the downloadable handout to follow along with as the calculation is completed. I usually project the problem on my classroom smart-board while solving the problem on the chalkboard as the students watch. This is the sample limiting-reactant problem:

If 25.4 g of mercury are placed with 10.5 g of oxygen and the reaction is allowed to go to completion, which is the limiting reactant and what mass of mercuric oxide will be produced?

Steps for solving sample problem:

  1. Balance the chemical equation:

    2 Hg + O2 → 2 HgO

  2. Write what is given and what needs to be calculated on the equation (moles go under the equation, all other units go above the equation).

    25.4g 10.5 g
    2 Hg + O2 → 2 HgO

  3. Draw a box containing two lines underneath any one of the reactants. This will serve as the chosen reactant for future calculations. Label the top line “Have” and the bottom line “Needed” (place an “X” on this line).

    Note: It doesn’t matter which reactant is chosen, but it will make solving the problem a little bit quicker if the box is placed under the limiting reactant. You may want to encourage students to draw the box under the reactant with the smaller quantity. Since there is less oxygen, I will guess that it will be the limiting reactant.
Csnov16samsa1
  1. Draw a path, using arrows, from the given data to the box under your chemical equation (shown above).

  2. Using the given quantity of the chosen reactant, calculate its number of moles and write the answer on the line labelled “Have.”

  3. Using the given quantity from the other reactant, calculate the number of moles needed for the chosen reactant and write the answer on the line labelled “Needed.”

    Note: It doesn’t matter whether you calculate the “Have” or “Needed” first.

    10.5 g O2 x 1 mole O2 = 0.328 moles O2 present
    32.0 g O2

    25.4 g Hg x 1 mol Hg x 1 mol O2 = 0.0633 moles O2 needed
    200.6 g Hg 2 mol Hg

    Now write those calculated values on the lines in the box.
Csnov16samsa2
  1. Determining the limiting reactant: If there is more reactant on the “Have” line than what is identified on the “Needed” line, then that reactant is in excess. If there is less of the chosen reactant than the quantity needed to completely react, then the chosen reactant is the limiting reactant.

Now students can clearly see that 0.0633 moles of oxygen are needed to make 25.4 g of mercury react, and that 0.328 moles of oxygen are available. Therefore, we have much more oxygen than what we need for the reaction. So the oxygen will still remain after the reaction is over, meaning it is in excess. The mercury will run out first and when it is gone the reaction will stop. Therefore, mercury is the limiting reactant.

Considerations for guiding the calculations

Notice that the number of dimensional analysis steps in each calculation above corresponds to the number of arrows on the student’s map. I use an analogy to help the students remember that, just like the mole (animal) lives under ground, the mole (unit) lives under the balanced equation. Since the vertical arrow starting at 10.5 g points downward to the “Have” box, it is indicating that the necessary calculation should convert grams of oxygen to moles. This represents how many moles of oxygen are present at the start of the reaction.

10.5 g O2 x 1 mole O2 = 0.328 moles O2 (have)
32.0 g O2
First arrow

The second calculation shows how many moles of oxygen are needed for all 25.4 g of mercury to react. Once again, the number of arrows on the student’s map corresponds to the number of steps in the dimensional analysis. The first arrow is vertical, starting at 25.4 g and points downward, through the equation, indicating a necessary conversion of mass of mercury to moles of mercury. The second arrow is horizontal, and is placed underneath Hg (moles of Hg) pointing toward “Needed” oxygen (moles of oxygen), indicating a mole-mole conversion is necessary.


25.4 g Hg x 1 mol Hg x 1 mol O2 = 0.0633 moles O2 (needed)
200.6 g Hg 2 mol Hg
First arrow Second arrow



Alternative solving option

If the student had chosen to put the box under the mercury, this is how the calculation would be completed:

Csnov16samsa3

25.4 g Hg x 1 mol Hg = 0.127 mol Hg (have)
200.6 g Hg
First arrow

10.5 g O2 x 1 mol O2 x 2 mol Hg = 0.656 mol Hg (needed)
32.0 g O2 1 mol O2
First arrow Second arrow

Csnov16samsa4

This calculation would lead the student to see that in order for 10.5 g of oxygen to react, it must come in contact with 0.656 moles of mercury. Since there are only 0.127 moles of mercury present, all of the oxygen will not react, so the oxygen is in excess. Since the oxygen is in excess, the limiting reactant is the mercury. In my experience, it seems easier for students who struggle with this concept to determine which reactant is in excess rather than which reactant is limiting. I teach students to use the following reasoning: if you have more than what you need, you have excess material.

Stoichiometry set-up method

The second part of the problem requires the student to determine the mass of mercuric oxide that will be produced from these reactants. I teach my students to write an “X” on the equation with the appropriate units needed for the product. From here, this method follows the "Stoichiometry Set-up Method" published in the September 2015 issue of Chemistry Solutions. Students draw a map to follow, starting with the determined limiting reactant and ending where the X is marked. If the problem had asked for moles of HgO, the X would be placed underneath the HgO. I prefer to have the student start the problem from the given value of the limiting reactant stated in the problem instead of from the calculated moles of limiting reactant. Doing so allows for a quick recheck of their initial calculation.

(Limiting Reactant)


The sequence of three arrows begins at 25.4 g Hg and ends at the mass of HgO produced, indicating three conversions are needed:


The sequence of three arrows begins at 25.4 g Hg and ends at the mass of HgO produced, indicating three conversions are needed:

25.4 g Hg x 1 mol Hg x 2 mol HgO x 216.6 g HgO = 27.4 g HgO
200.6 g Hg 2 mol Hg 1 mol HgO
First arrow Second arrow Third arrow

The limiting reactant problem-solving map is particularly helpful for students when they don’t know what to do next during the calculation. Placing a box under a reactant is a great way to quickly see how much of it is present for the reaction and how much of it is needed. By highlighting those results with a box, it lets the student see only the final results of a sometimes overwhelming (and perhaps distracting) calculation, for a quick reference of how much of a reactant is “had” and how much of it is “needed.”

Another benefit to this method is that it allows the student to determine the amount of excess reactant that still remains after the reaction has completed. This is a commonly asked follow-up question to a limiting reactant calculation, and the box created under the equation serves as a very good guide to its solution. By subtracting the moles “had” from the moles “needed” of the excess reactant, the moles of excess reactant that remain after the reaction is completed can be quickly determined.

To illustrate this in the above problem: It’s already determined that oxygen is in excess from the initial calculation, so subtract the “Have” from the “Needed” mole values:

0.328 mol (O2 have) – 0.0633 mol (O2 needed) = 0.265 mol (O2 remains)

In addition, since the original masses of the reactants were in grams, converting the remaining quantity of moles to grams is simple:

0.265 mol O2 x 32.0 g O2 = 8.48 g O2 remains
1 mol O2

Sometimes seeing all of the numbers involved in one of these limiting reactant problems causes students to “not see the forest for the trees.” Students who are struggling with the process of stoichiometry sometimes tend to forget why they are taking the steps they are following. Having the completed box underneath the equation allows the student to get a quick summary of how much of the reactant is present versus how much of it is needed. This condensed short-story version of the long novel sometimes enables the student to better focus on those two most important parts of the limiting reactant calculation.