Chemistry · Grades 10–12
Limiting amounts in a model
Divide each reactant amount by its equation coefficient. The smaller reaction extent determines the possible product.
Rule and assumptions
Divide each reactant amount by its equation coefficient. The smaller reaction extent determines the possible product.
A + 2B → C; n(C) = min(n(A), n(B)/2)
Worked examples
For the abstract reaction A + 2B → C, 43 mol A and 44 mol B are available. Assume complete consumption of the limiting reactant. Find the maximum amount of C. Round to 3 decimal places if needed. Answer in mol.
A + 2B → C; n(C) = min(n(A), n(B)/2) min(43, 44/2) = 22
22 mol
For the abstract reaction A + 2B → C, 50 mol A and 40 mol B are available. Assume complete consumption of the limiting reactant. Find the maximum amount of C. Round to 3 decimal places if needed. Answer in mol.
A + 2B → C; n(C) = min(n(A), n(B)/2) min(50, 40/2) = 20
20 mol
Try two questions
For the abstract reaction A + 2B → C, 7 mol A and 60 mol B are available. Assume complete consumption of the limiting reactant. Find the maximum amount of C. Round to 3 decimal places if needed. Answer in mol.
For the abstract reaction A + 2B → C, 49 mol A and 12 mol B are available. Assume complete consumption of the limiting reactant. Find the maximum amount of C. Round to 3 decimal places if needed. Answer in mol.
Answers and reasoning
Answers and marking notes
- 7 mol
A + 2B → C; n(C) = min(n(A), n(B)/2) min(7, 60/2) = 7
- 6 mol
A + 2B → C; n(C) = min(n(A), n(B)/2) min(49, 12/2) = 6
Further reading and review references
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