Materials & Prep
Prepare a one-page practice sheet with the worked reaction and independent problems below. Provide calculators, periodic tables, and the following atomic masses: H 1.008, O 16.00, Al 26.98, Cl 35.45, Br 79.90. Write 2H₂ + O₂ → 2H₂O on the board. Prepare six H2 particle cards and two O2 particle cards for the opening.
Opening prediction
Display six H2 cards and two O2 cards without initially naming them as moles. Tell students that one complete reaction requires two H2 cards and one O2 card. Ask: “How many complete reaction groups can you build, and which card runs out first?” Students arrange or sketch the groups, then commit to a prediction individually before discussing. The class should find two complete groups, with H2 left over and O2 used up.
ConnectExplain that students just identified a limiting reactant by comparing how many complete reaction groups each substance could support. Today, the particles will be counted in moles rather than one at a time. Emphasize that the smaller mass is not automatically limiting. The substance that supports fewer complete reaction sets is limiting.
Build the mole method
ExplainA balanced equation compares particles in mole ratios, not masses. For 2H₂ + O₂ → 2H₂O, 2 moles of H₂ are required for every 1 mole of O₂. Convert each starting mass to moles, then compare how much reaction each reactant can support by dividing available moles by its coefficient.
Worked example
10.0 g H₂ ÷ 2.016 g/mol = 4.96 mol H₂ 32.0 g O₂ ÷ 32.00 g/mol = 1.00 mol O₂
H₂: 4.96 mol ÷ 2 = 2.48 possible reaction units O₂: 1.00 mol ÷ 1 = 1.00 possible reaction unit
Think aloudI cannot compare 4.96 mol H₂ directly with 1.00 mol O₂ because the equation does not use them one-to-one. After accounting for the coefficients, O₂ supports fewer reaction units, so O₂ is the limiting reactant. The common error is stopping after noticing that 10.0 g is less than 32.0 g. The masses cannot be compared directly.
Continue the calculation from the limiting reactant: 1.00 mol O₂ × (2 mol H₂O ÷ 1 mol O₂) = 2.00 mol H₂O. Then 2.00 mol H₂O × 18.016 g/mol = 36.0 g H₂O.
ExpectedO₂ is limiting, and 36.0 g H₂O forms. The 32.0 g sample is limiting even though it has more mass.
Check for understandingStudents answer on paper: If 3.00 mol H₂ and 2.00 mol O₂ begin the same reaction, which supports fewer reaction units? Expected answer: H₂, because 3.00 ÷ 2 = 1.50 while 2.00 ÷ 1 = 2.00. Ask one student to explain why the smaller quotient, not the smaller mass, identifies the limiting reactant.
Guided practice
ModelUse 2Al + 3Cl₂ → 2AlCl₃ with 5.40 g Al and 14.2 g Cl₂. Give students this organizer: mass → moles → divide by coefficient → smaller value identifies limiting reactant → use limiting reactant to find product.
Have students complete the first two conversions, then finish with a partner:
5.40 g Al ÷ 26.98 g/mol = 0.200 mol Al 14.2 g Cl₂ ÷ 70.90 g/mol = 0.200 mol Cl₂
Al: 0.200 ÷ 2 = 0.100 reaction units Cl₂: 0.200 ÷ 3 = 0.0667 reaction units
ExpectedCl₂ is limiting. Product calculation: 0.200 mol Cl₂ × (2 mol AlCl₃ ÷ 3 mol Cl₂) = 0.133 mol AlCl₃. Then 0.133 mol × 133.33 g/mol = 17.8 g AlCl₃.
CirculateLook for students who compare 5.40 g and 14.2 g directly or divide by the wrong coefficient. Prompt them to point to the coefficient for each reactant in the balanced equation before calculating.
Independent analysis
Students solve both problems individually. Require four labeled lines for each: moles of each reactant, reaction units for each, limiting reactant, and grams of product.
Problem 1: N₂ + 3H₂ → 2NH₃. Starting amounts: 14.0 g N₂ and 2.00 g H₂. Use N = 14.01 and H = 1.008.
Problem 2: 2CO + O₂ → 2CO₂. Starting amounts: 56.0 g CO and 32.0 g O₂. Use C = 12.01 and O = 16.00.
Expected for Problem 1: 14.0 ÷ 28.02 = 0.500 mol N₂; 2.00 ÷ 2.016 = 0.992 mol H₂. Reaction units are 0.500 for N₂ and 0.331 for H₂, so H₂ is limiting. Product is 0.992 × (2 ÷ 3) = 0.661 mol NH₃, or 0.661 × 17.03 = 11.3 g NH₃.
Expected for Problem 2: 56.0 ÷ 28.01 = 2.00 mol CO; 32.0 ÷ 32.00 = 1.00 mol O₂. Reaction units are 1.00 for CO and 1.00 for O₂, so the reactants are in exact stoichiometric proportions. Product is 2.00 mol CO₂, or 2.00 × 44.01 = 88.0 g CO₂. Students should state that neither reactant is in excess.
ScaffoldStudents who need support may use the organizer and a sentence frame: “The limiting reactant is because supports reaction units, which is fewer than .” Accept calculator use, but require students to label units at every step.
ExtensionStudents who finish early calculate the leftover reactant in Problem 1. The 0.500 mol N₂ would require 0.500 × 3 = 1.50 mol H₂, but only 0.992 mol H₂ is available, so H₂ is the limiting reactant. N₂ consumed is 0.992 ÷ 3 = 0.331 mol. Leftover N₂ is 0.500 − 0.331 = 0.169 mol, or 4.73 g. Students explain why the excess reactant remains.
Closing check
Exit ticketFor 2Al + 3Br₂ → 2AlBr₃, a reaction begins with 5.40 g Al and 24.0 g Br₂. Use Al = 26.98 and Br = 79.90. Identify the limiting reactant and calculate the mass of AlBr₃ formed. Show the mole conversion and stoichiometric comparison.
Expected5.40 ÷ 26.98 = 0.200 mol Al; 24.0 ÷ 159.80 = 0.150 mol Br₂. Reaction units are 0.100 for Al and 0.0500 for Br₂, so Br₂ is limiting. Product is 0.150 × (2 ÷ 3) = 0.100 mol AlBr₃; 0.100 × 266.68 = 26.7 g AlBr₃. Collect responses and check that students used moles and coefficients rather than comparing the two masses.