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Stoichiometry and Limiting Reactant

By Chuck Roser, Retired Chemistry Instructor
North Carolina School of Science and Mathematics

Updated October 2016

Limiting ReactantNational Science Education Standards

This activity is appropriate for high school students and addresses the following National Science Education Standards for grades 9–12:

  1. Science as Inquiry: Understandings About Scientific Inquiry
  2. Physical Science: Chemical Reactions

Objectives

  1. To gain understanding of the concept of a limiting reactant in a chemical reaction
  2. To determine the combining ratios for reactions of calcium chloride with sodium oxalate and sodium phosphate
  3. To write a balanced equation for each reaction

Introduction

A balanced chemical equation specifies the mole ratio in which the reactants in a chemical reaction combine to form the products. You derive these combining ratios from the coefficients in the balanced equation: They may or may not be a 1:1 ratio. Consider the example of hydrogen and oxygen reacting to form water:

2 H2 (g) + O2 (g) →   2 H2O (l)

The coefficients indicate 2 molecules of hydrogen combine with a molecule of oxygen to form 2 molecules of water. You may also interpret the coefficients in a balanced equation as the mole ratios between the reactants to form the products. Two moles of hydrogen and 1 mole of oxygen react to form 2 moles of water. If 4 moles of hydrogen and 1 mole of oxygen are reacted, will the reactants completely convert to product or will 1 of the reactants limit the amount of water formed? Examining the combining ratio of 2 moles of hydrogen required for each mole of oxygen, the oxygen is consumed before the hydrogen. Two moles of hydrogen remain unreacted. The 1 mole of oxygen limits the amount of product formed. The oxygen is the limiting reactant for this reaction. The mole ratio of the reactants and the actual amount of the compounds put into the reaction determine the amount of product formed. The maximum amount of product forms when the reactants are in the same mole ratio as the coefficients in the balanced chemical equation since both reactants are completely used.

This lab uses the method of continuous variations to determine the mole ratio between the reactants. You use equal molar concentration solutions so that the volume ratio of the solutions is the same as the mole ratio. The total volume of the solutions in each mixture is held constant so the total number of moles is constant. If the reactants are not in the same mole ratio as the coefficients in the balanced equation, then 1 reactant is in excess, some of that reactant remains at the end of the reaction, and less product forms. By varying the volume ratio, and thus the mole ratio, and observing the amount of product formed, you determine the coefficients in the balanced equation. You can then write a balanced equation for the reaction.

Materials (for a class of 20 students working in pairs)

  • 20 Safety Glasses (plus a pair for the instructor)
  • 10 24-Well Plates
  • 10 Dropper Bottles or Beakers (with dropping pipets labeled 0.10 M CaCl2)
  • 10 Dropper Bottles or Beakers (with dropping pipets labeled 0.10 M Na2C2O4)
  • 10 Dropper Bottles or Beakers (with dropping pipets labeled 0.10 M Na3PO4)

Note: Dropping bottles or dropping pipets must be the same size so that they deliver the same size drops. The total volume of each reactant and the total moles of each reactant will also be the same.

Safety

  1. Students and instructor should wear safety glasses during data collection.
  2. Students should follow the instructor’s directions for cleaning equipment and disposing of waste chemicals.

Procedure I: calcium chloride and sodium oxalate

Well Number 1 2 3 4 5 6 7 8 9
Drops of H2O 4 4 4 4 4 4 4 4 4
Drops of CaCl2 1 2 3 4 5 6 7 8 9
Drops of Na2C2O4 9 8 7 6 5 4 3 2 1
Figure 1  Distribution of chemicals for Procedure I.

  1. Place 4 drops of water in each of 9 wells in the 24-well plate. (See Fig. 1.)
  2. Add 1 drop of 0.10 M calcium chloride (CaCl2) to well 1, 2 drops to well 2, 3 drops to well 3, etc., until you add 9 drops to well 9. Try to keep the same drop size.
  3. Add 9 drops of 0.10 M sodium oxalate (Na2C2O4) to well 1, 8 drops to well 2, 7 drops to well 3, etc., until you add 1 drop to well 9. Try to keep the same drop size. Note: To prevent contamination of the solution in the pipet, do not touch the pipet to the solutions in the well plate.
  4. Mix the contents of each filled well by gently swirling the plate, being careful not to spill any of the contents. Allow about 5 min for precipitate to settle.
  5. Observe the solids in each of the wells and visually determine which well has the most precipitate. If 2 wells are difficult to rank, redo those. Recall the drop ratio of CaCl2 to Na2C2O4 for this well.
  6. Since the reactant solutions have the same concentrations (molarities), the volume ratio of the reactants is the same as the mole ratio. On this drop ratio basis, determine ratio of the coefficients of the reactants.
  7. Follow your instructor’s directions for cleaning the equipment and disposing of the waste chemicals.

Procedure II: calcium chloride and sodium phosphate


Well Number 1 2 3 4 5 6 7 8 9
Drops of H2O 4 4 4 4 4 4 4 4 4
Drops of CaCl2 1 2 3 4 5 6 7 8 9
Drops of Na3PO4 9 8 7 6 5 4 3 2 1
Figure 2  Distribution of chemicals for Procedure II.

  1. Repeat steps 1 through 5 in the procedure above using 0.10 M calcium chloride (CaCl2) and 0.10 M sodium phosphate (Na3PO4). (See Fig. 2.)
  2. Since the reactant solutions have the same concentrations (molarities), the volume ratio of the reactants is the same as the mole ratio. On this drop ratio basis, determine ratio of the coefficients of the reactants.
  3. Follow your instructor’s directions for cleaning the equipment and disposing of the waste chemicals.

Instructor’s notes

  1. The solutions are 0.10 M in all cases and can be prepared as follows:

    0.10 M CaCl2 • 2 H2O: 1.47 g/100 mL

    0.10 M Na2C2O4 : 1.34 g/100 mL

    0.10 M Na3PO4 • 12 H2O : 3.8 g/100 mL

    24-well plates


  2. In Procedure I, the maximum amount of precipitate should be in well 5 since the CaCl2 and Na2C2O4 react in a 1:1 ratio. Before well 5, the CaCl2 is limiting. After well 5, the Na2C2O4 is limiting.


    CaCl2 (aq) + Na2C2O4 (aq)  →  CaC2O4 (s) + 2 NaCl (aq)


    In Procedure II, the maximum amount of precipitate should be in well 6 since the CaCl2 and Na3PO4 react in a 1.5:1 or 3:2 ratio. Before well 5, the CaCl2 is limiting. After well 5, the Na3PO4 is limiting.


    3 CaCl2 (aq) + 2 Na3PO4 (aq)  →  Ca3(PO4) 2 (s) + 6 NaCl (aq)

Discussion of results

Once the students complete the lab, a guided discussion of their results helps reinforce the concept of a limiting reactant. Have the students reach a consensus on which well in each reaction produced the maximum amount of precipitate. In these wells, the reactants should be in the same ratio as the coefficients in the balanced equation. Ask which reactant is limiting and which reactant is in excess in the other wells. If the students are familiar with double-displacement reaction, they can use the ratio between the reactants to write a complete balanced equation for the reaction. If they are familiar with solubility rules, have them predict which compound is the precipitate and add physical states to their reactions. A discussion of which variables are held constant in the lab, why they are held constant, and sources of error helps them understand experimental design and evaluation of results. Limiting reactants is often a difficult concept for students to understand, but allowing students to perform simple reactions helps reinforce the concept.

In terms of practical applications, industrial chemists run reactions to produce a compound to sell. Engage the students in a discussion about why it might be economical to have the reactants in the correct mole ratio, how it might be easier to separate and purify the product without excess reactant present, and the possibility of recycling any excess reactant. Ask students for other examples of a limiting reactant or resource in their lives.

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