Mole concept & limiting reagents

A balanced reaction gives ratios of amounts in moles. It does not generally give the same ratios of masses. Convert each given mass using n = m/M, then divide each reactant amount by its stoichiometric coefficient. The smallest result identifies the limiting reactant.

A worked example

For 2Al + 3Cl₂ → 2AlCl₃, compare n(Al)/2 with n(Cl₂)/3. If you have 0.400 mol Al and 0.450 mol Cl₂, the values are 0.200 and 0.150. Chlorine limits the reaction. It produces 2 × 0.150 = 0.300 mol AlCl₃.

A common mistake

Do not apply percentage yield until you have calculated the theoretical product amount from the limiting reagent.

Try the idea

These original practice questions also appear in the foundation mock. Attempt them before opening the explanation.

Question 1 · Mole conceptSingle correct · +3 / −1

A 5.00 g mixture of Na₂CO₃ and NaHCO₃ is heated until only Na₂CO₃ remains. Its mass decreases by 0.620 g. What was the mass of NaHCO₃ in the original mixture? Use Na = 23, C = 12, O = 16, H = 1.

Select one answer.A. 0.840 gB. 1.68 gC. 2.48 gD. 3.36 g

Question 2 · Mole conceptSingle correct · +3 / −1

An oxide contains 70.0% iron and 30.0% oxygen by mass. With Fe = 56 and O = 16, its empirical formula is:

Select one answer.A. FeOB. Fe₂O₃C. Fe₃O₄D. FeO₂

Question 3 · Mole conceptSingle correct · +3 / −1

For 2Al + 3Cl₂ → 2AlCl₃, 0.400 mol Al is mixed with 0.450 mol Cl₂. Assuming complete reaction, how much Al remains?

Select one answer.A. 0 molB. 0.050 molC. 0.100 molD. 0.150 mol

Question 47 · Practical reasoningSingle correct · +3 / −1

A student prepares 250.0 mL of 0.1000 M Na₂CO₃ using anhydrous solid. What mass is required? M(Na₂CO₃) = 106.0 g mol⁻¹.

Select one answer.A. 0.2650 gB. 1.060 gC. 2.650 gD. 26.50 g

What to do next

Explain your answer aloud, then find a related question in an official paper. If you can only follow a worked solution but cannot start a fresh question, revisit the central idea and try again without the notes.