Predict yields, and find what runs out.
A balanced equation is an atomic recipe. In the lab you rarely mix the reactants in the exact ratio the recipe needs. One reactant runs out first and stops the reaction. A BCA (Before-Change-After) table tracks what is used up and what forms. Work in moles and coefficients, and you can predict how much product forms and what is left over.
Core Claims
- Mole ratios: The coefficients of a balanced equation give the exact ratio of moles between reactants and products.
- BCA tables: A Before-Change-After table tracks the moles of every substance. The Change row follows the coefficients.
- Limiting reactant: The reactant that runs out first stops the reaction and sets the maximum yield. The other reactants are in excess.
Reaction yield map
Retrieval Checklist
- Solve a stoichiometry problem with a BCA table.
- Name the limiting and excess reactants from starting moles.
- Calculate the moles left over when the reaction ends.
Coefficients are scaling factors.
A sandwich recipe: 1 slice of cheese + 2 slices of bread → 1 sandwich. The ratio of cheese to bread is 1:2. With 5 slices of cheese and 6 slices of bread, you can make only 3 sandwiches. The bread runs out first and leaves 2 slices of cheese.
In chemistry, the coefficients of a balanced equation give the same kind of proportion, measured in moles. Consider the combustion of methane gas:
This equation shows that 1 mole of methane reacts with 2 moles of oxygen. It makes 1 mole of carbon dioxide and 2 moles of water. This is the mole ratio. Reacting 3 moles of CH4 takes 6 moles of O2. It makes 3 moles of CO2 and 6 moles of H2O.
Reading a coefficient as grams. The 2 in front of O2 means 2 moles of O2 for every 1 mole of CH4.
The accounting table of chemistry.
To find a reaction yield, set up a BCA table (Before, Change, After). A BCA table tracks every substance at the same time:
- Before (B): The starting moles of the reactants before the reaction begins. The products start at 0.
- Change (C): The moles used up (negative) or made (positive). The values in this row must scale in the exact ratio of the equation's coefficients.
- After (A): The moles left when the reaction ends (Before + Change).
Here is a balanced BCA table. 2.0 moles of CH4 and 4.0 moles of O2 are mixed in the exact 1:2 ratio. Both reactants are fully consumed:
| Row | CH4 | + 2 O2 | → CO2 | + 2 H2O |
|---|---|---|---|---|
| Before (B) | 2.0 mol | 4.0 mol | 0 mol | 0 mol |
| Change (C) | -- | -- | -- | -- |
| After (A) | -- | -- | -- | -- |
| Row | CH4 | + 2 O2 | → CO2 | + 2 H2O |
|---|---|---|---|---|
| Before (B) | 2.0 mol | 4.0 mol | 0 mol | 0 mol |
| Change (C) | −2.0 mol | −4.0 mol | +2.0 mol | +4.0 mol |
| After (A) | -- | -- | -- | -- |
| Row | CH4 | + 2 O2 | → CO2 | + 2 H2O |
|---|---|---|---|---|
| Before (B) | 2.0 mol | 4.0 mol | 0 mol | 0 mol |
| Change (C) | −2.0 mol | −4.0 mol | +2.0 mol | +4.0 mol |
| After (A) | 0 mol | 0 mol | 2.0 mol | 4.0 mol |
Notice that the Change row values (−2.0, −4.0, +2.0, +4.0) match the 1:2:1:2 coefficient ratio of the equation. Reactants are consumed and products are built in locked step.
Filling the Change row with equal numbers. The row follows the coefficients: −2.0 mol CH4 goes with −4.0 mol O2, not −2.0 mol.
The limiting reactant sets the limit.
What if you mix 2.0 moles of CH4 with 5.0 moles of O2? There is not enough methane to react with all of the oxygen. Methane runs out first.
The reactant that is completely consumed first is the limiting reactant. It limits how much product can form. The reactant that remains after the reaction halts is the excess reactant.
To find the limiting reactant with a BCA table, test which reactant would hit 0 first if it were fully consumed. The one with the smaller Change row multiplier is the limiting reactant. When it hits 0 in the After row, the reaction stops. The excess reactant stays in the container unchanged.
Picking the reactant with the fewest moles. In Question 1 below, 3.0 mol CH4 is less than 4.0 mol O2, yet O2 runs out first, because each CH4 needs 2 O2.
The fewest pieces is not always the limiting reactant.
Many students think the reactant with the smallest amount always runs out first. The s'mores test shows why that is wrong. Each s'more needs 2 graham crackers + 1 marshmallow + 1 chocolate square. Having the fewest marshmallows does not make them limiting. You have to check the recipe ratio.
Which ingredient runs out first?
The explanation appears after you predict.
- Graham cracker
- Marshmallow
- Chocolate square
- The limiting ingredient's bar
- Raise the marshmallows to 12. Look for: the limiting ingredient does not change, because the crackers still make only 5 s'mores.
- Lower the crackers to 6, with 6 marshmallows and 8 chocolate squares. Look for: the limiting ingredient changes, even though crackers are still the largest count.
From s'mores to chemistry: the BCA ratio method
Every limiting reactant problem follows the same logic. Divide each reactant's available moles by its coefficient. The smallest quotient identifies the limiting reactant.
The limiting reactant has the smallest value of: available moles ÷ coefficient.
The s'mores check: crackers ÷ 2 = 5, marshmallows ÷ 1 = 6, chocolate ÷ 1 = 8. The crackers are limiting, because 5 is the smallest.
Slide the progress. Watch the atoms rearrange.
Set the starting moles of methane (CH4) and oxygen (O2). Slide the reaction progress to watch the molecules burn, and read the moles in the BCA table as they change. One drawn molecule stands for half a mole.
How much of each substance is left when the reaction stops?
| Row | CH4 | + 2 O2 | → CO2 | + 2 H2O |
|---|---|---|---|---|
| Before (B) | – | – | 0.00 | 0.00 |
| Change (C) | – | – | – | – |
| After (A) | – | – | – | – |
The explanation appears after you predict.
- Carbon atom
- Hydrogen atom
- Oxygen atom
- Flame: a reaction in progress
- Load 2.0 mol CH4 with 5.0 mol O2, then press Record. Look for: CH4 reaches 0, and 1.00 mol of O2 is left over.
- Load 3.0 mol CH4 with 4.0 mol O2, then press Record. Look for: now O2 reaches 0, and 1.00 mol of CH4 is left over.
- Load 2.0 mol CH4 with 4.0 mol O2, then press Record. Look for: both reach 0, because the amounts are in the exact 1:2 ratio.
Use your evidence
Predict the reaction yield.
Fill the blanks from memory.
Type each blank from memory, then press Check. Press Show answers after your first check if you are stuck.
The study of the amounts of reactants and products in a reaction is . The reactant that is used up first and limits the product is the . The reactant left over when the reaction stops is the . In a BCA table, the letter B stands for . The Change row of a BCA table must match the ratio of the equation's .
Explain why the Change row must match the equation's coefficients, even when the Before amounts differ.
Writing it out forces your brain to organize the idea. Write your answer first, then peek and compare.
The Before row holds whatever amounts you happened to mix. But molecules react in fixed proportions set by the balanced equation. Each CH4 molecule needs 2 O2 molecules and makes 1 CO2 and 2 H2O. So the particles used up and made always follow the coefficient ratio. The Change row tracks that, so it must match the coefficients exactly.
Write your answer first. Then grade yourself.
Give yourself 1 mark for each idea you wrote down.
A mixture holds 1.50 moles of methane (CH4) and 2.00 moles of oxygen (O2). They react: CH4 + 2 O2 → CO2 + 2 H2O. Name the limiting reactant. Calculate the maximum moles of H2O formed. Find the moles of excess reactant left over. Show all work.
- Setup (1 mark): You list the starting amounts: 1.50 mol CH4, 2.00 mol O2 and 0 mol H2O.
- Limiting reactant (1 mark): You name O2 and show that 1.50 mol CH4 needs 3.00 mol O2.
- Yield (1 mark): You find 2.00 mol H2O, from 2.00 mol O2 in a 2:2 ratio.
- Excess (1 mark): You find 0.50 mol CH4 left over (1.50 mol minus 1.00 mol reacted).
Fertilizer depends on stoichiometry. The Haber-Bosch process makes ammonia (N₂ + 3 H₂ → 2 NH₃), and ammonia fertilizer feeds half the world's population. Mixing the reactants in the exact 1:3 ratio wastes none of them.
Adaptive Competency Practice: HS-PS1-7
Solve stoichiometry ratios, limiting reactant scenarios, and theoretical percent yields with adaptive support. Gain instant feedback and an authenticated completion seal.