To solve limiting reactant practice problems, convert each reactant to moles, divide by its coefficient in the balanced equation, and compare the results. The smaller result identifies the reactant that limits the theoretical product amount. Comparing starting grams, or even starting moles without using coefficients, can give the wrong answer.
The eight exercises below move from mole quantities to masses, leftover reactants, solution concentrations, and incomplete information. They are original classroom calculations, not instructions for performing reactions. Assume the stated reaction is the only reaction and proceeds to completion unless the question says otherwise. Keep extra digits during calculations and round the final answer using the supplied data.
The setup that keeps quantities straight
Start with a balanced equation. Coefficients tell you the proportions in which particles or moles react. They do not tell you the proportions of their masses. If balancing is the sticking point, work through these balancing chemical equations exercises before adding quantities.
For a reaction written aA + bB → cC, calculate n(A)/a and n(B)/b. Call the smaller value the reaction amount. Multiply it by c to find the theoretical moles of C. An equivalent method calculates the possible amount of the same product from each reactant separately, then chooses the smaller amount. OpenStax explains both the limiting-reactant idea and the product-comparison method.
Use three lines for every reactant: starting amount, amount consumed, and amount remaining. This small inventory catches a common mistake: correctly identifying the limiting reactant but using the entire starting amount of the excess reactant in the next calculation.
Keep the chemical formula beside each mole value. Writing only “0.25 mol” makes it easy to confuse moles of reactant with moles of product. In a conversion factor, put the substance you want above the substance you are converting so the starting label cancels.
Problems 1 and 2: Start with moles
1. Which reactant runs out?
For N2 + 3H2 → 2NH3, the starting quantities are 2.00 mol N2 and 4.50 mol H2. Identify the limiting reactant and calculate theoretical ammonia and leftover nitrogen.
Answer: Nitrogen gives 2.00/1 = 2.00 mol of reaction amount. Hydrogen gives 4.50/3 = 1.50 mol. Hydrogen limits the reaction. The theoretical ammonia amount is 1.50 × 2 = 3.00 mol NH3.
The reaction consumes 1.50 mol N2, leaving 2.00 – 1.50 = 0.50 mol N2. All 4.50 mol H2 is consumed in this idealized calculation. Check the hydrogen inventory: 3.00 mol ammonia contains 9.00 mol hydrogen atoms, exactly the amount supplied by 4.50 mol hydrogen molecules.
Tempting mistake: Choosing nitrogen because 2.00 is smaller than 4.50. That comparison ignores the three hydrogen molecules required for every nitrogen molecule.
2. Can the larger mole amount be limiting?
For 4Al + 3O2 → 2Al2O3, use 0.800 mol Al and 0.700 mol O2. Find the limiting reactant, product amount, and oxygen remaining.
Answer: Aluminum gives 0.800/4 = 0.200; oxygen gives 0.700/3 = 0.2333. Aluminum is limiting, even though its starting mole quantity is larger. Product is 0.200 × 2 = 0.400 mol Al2O3.
Oxygen consumed is 0.200 × 3 = 0.600 mol, so 0.100 mol remains. Your inventory should never show a negative remainder. A negative result often means that you calculated consumption from the excess reactant instead of from the limiting amount.
Problems 3 and 4: Convert mass and track leftovers
3. Equal grams do not mean equal reacting amounts
For 2H2 + O2 → 2H2O, start with 4.00 g of each reactant. Use molar masses H2 = 2.016 g/mol, O2 = 32.00 g/mol, and H2O = 18.016 g/mol. Find the theoretical water mass.
Answer: Hydrogen: 4.00/2.016 = 1.9841 mol, then 1.9841/2 = 0.99206. Oxygen: 4.00/32.00 = 0.125 mol, then 0.125/1 = 0.125. Oxygen limits the reaction.
Water is 0.125 × 2 = 0.250 mol. Its mass is 0.250 × 18.016 = 4.504 g, reported as 4.50 g. The product can weigh more than either starting reactant alone because it includes atoms from both.
For an extra check, hydrogen consumed is 0.250 × 2.016 = 0.504 g. The unrounded inventory is 4.504 g water plus 3.496 g hydrogen remaining, totaling the original 8.00 g. Do not round each intermediate line before checking the mass balance.
4. How much excess reactant remains?
For 2Mg + O2 → 2MgO, begin with 6.00 g Mg and 5.00 g O2. Use Mg = 24.31 g/mol and O2 = 32.00 g/mol. Find the oxygen left over.
Answer: Magnesium gives (6.00/24.31)/2 = 0.123406 reaction amount. Oxygen gives 5.00/32.00 = 0.15625. Magnesium is limiting. Oxygen consumed is 0.123406 mol, or 3.94899 g.
Oxygen remaining is 5.00 – 3.94899 = 1.05101 g, reported as 1.05 g. Subtract oxygen consumed from oxygen supplied. Subtracting grams of magnesium from grams of oxygen has no stoichiometric meaning, even though both values use grams.
Problems 5 and 6: Special cases
5. What if both ratios are equal?
For 2CO + O2 → 2CO2, a hypothetical mixture contains 0.600 mol CO and 0.300 mol O2. Is either reactant in excess?
Answer: Both normalized amounts equal 0.300. The reactants are supplied in the exact stoichiometric ratio at the stated precision. Neither is in excess; both are exhausted together in the idealized model. The theoretical product is 0.600 mol CO2.
Some courses describe both as limiting in this special case. The important conclusion is the inventory: no reactant remains under the assumptions. Do not force a unique limiting choice by rounding one ratio differently from the other.
6. A solution problem with milliliters
For AgNO3 + NaCl → AgCl + NaNO3, the given solutions are 30.0 mL of 0.200 mol/L AgNO3 and 40.0 mL of 0.100 mol/L NaCl. Find theoretical moles of AgCl. Treat precipitation as complete.
Answer: Convert volumes to liters. AgNO3: 0.0300 L × 0.200 mol/L = 0.00600 mol. NaCl: 0.0400 L × 0.100 mol/L = 0.00400 mol. All coefficients are one, so NaCl is limiting and 0.00400 mol AgCl forms.
The remaining silver-ion amount is 0.00200 mol. You do not need the combined solution volume to identify the limiting amount. You would need a final-volume assumption if asked for the concentration of an ion after mixing. Read the requested unit before doing extra calculations.
Problems 7 and 8: Yield and missing information
7. Connect theoretical and actual yield
A separate calculation gives a theoretical yield of 12.0 g. The actual collected product is 9.30 g. What is the percent yield, and does this tell you how much reactant remains?
Answer: Percent yield is (9.30/12.0) × 100 = 77.5%. It does not, by itself, tell you the leftover reactant quantity. Product might have been lost during collection, a competing reaction might have occurred, or conversion might have been incomplete.
Do not automatically multiply every theoretical reactant remainder by 77.5%. You need information about what caused the yield difference. OpenStax distinguishes actual yield from theoretical yield and discusses several reasons they differ.
8. Is there enough information?
A question gives 2.00 g H2 and an unspecified quantity of O2 for water formation. Can you identify the limiting reactant?
Answer: No. You need the oxygen amount or a statement that oxygen is in excess. If oxygen is explicitly in excess, hydrogen limits the theoretical water amount. Without that condition, the same hydrogen mass could be paired with too little oxygen, exactly enough oxygen, or extra oxygen.
A good written response names the missing information. It does not invent an oxygen quantity just to finish a familiar calculation. Check whether a diagram, table, or previous part of the question supplies that quantity before declaring the problem incomplete.
Turn your corrections into useful notes
On a fresh page, redo one mole problem, one mass problem, and the solution problem without looking at the answers. For each, explain why the other reactant is not limiting. Then compare your inventory with the worked solution and mark the first line where your reasoning changed.
Keep errors specific: coefficients ignored, milliliters not converted, product calculated from excess reactant, or unlike substances subtracted. Compare your corrected method with relevant chemistry course notes or browse student study documents. Check that any shared sheet uses a balanced equation and labels units.
Students should upload their own notes to help other students. Share original explanations and clearly labeled corrections rather than an instructor’s restricted answer key. A short, accurate inventory example can help someone who understands the formula but keeps choosing the wrong starting quantity.
Frequently asked questions
Is the smallest mass always limiting?
No. Convert mass to moles and account for coefficients. Problem 3 has equal masses but unequal reacting capacities.
Why divide moles by the coefficient?
It expresses each supply on the same reaction scale. The smaller supply on that scale determines how far the stated reaction can proceed.
Can I compare product amounts instead?
Yes. Calculate the same product from each reactant independently. The smaller predicted product amount is the theoretical maximum for the mixture.