Stoichiometry & BCA Tables

Predict yields, and find what runs out.

A balanced chemical equation is a microscopic recipe. In the laboratory, however, we rarely mix reactants in the exact ratio they require. One reactant will inevitably run out first, bringing the entire reaction to a halt. To keep track of reactant consumption and product generation, we use a simple framework called a BCA (Before-Change-After) table. By focusing on moles and stoichiometric coefficients, you can predict exactly how much product will form and what will be left over.

Alignment

HS-PS1-7 Calculate stoichiometric yields and identify limiting/excess reactants (PS1-7.3, .4 targets).

Objective

Complete a BCA table for a chemical reaction, identify the limiting reactant, and calculate product yield.

Scope

Mole-to-mole stoichiometry, BCA tables, limiting and excess reactants, conservation of atoms.

Core Claims

Mole Ratios: Balanced equation coefficients act as chemical recipes, establishing the exact proportional ratio required between reactants and products.
BCA Tables: Before-Change-After tables are used to track chemical reactions in moles (Change values are scaled directly to the coefficients).
Limiting Reactant: The reactant that runs out first, stopping the reaction and dictating the maximum yield of products. Excess reactants remain partially unreacted.

Reaction Yield Map

Retrieval Checklist

Solve stoichiometry problems using BCA tables.
Identify limiting and excess reactants from starting moles.
Calculate excess moles remaining after reaction completion.

01Proportion

Coefficients are scaling factors.

Think of a recipe for a sandwich: 1 slice of cheese + 2 slices of bread → 1 sandwich. The ratio of cheese to bread is strictly 1:2. If you have 5 slices of cheese and 6 slices of bread, you cannot make 5 sandwiches—you will run out of bread after making 3 sandwiches, leaving 2 slices of cheese unused.

In chemistry, the coefficients of a balanced equation represent this exact same proportion, but measured in moles. Consider the combustion of methane gas:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

This equation tells us that 1 mole of methane requires 2 moles of oxygen gas to react completely, producing 1 mole of carbon dioxide and 2 moles of water. This is our mole ratio. It is a scaling multiplier. If you react 3 moles of CH₄, you will need exactly 6 moles of O₂, and you will produce 3 moles of CO₂ and 6 moles of H₂O.

02Before-Change-After

The accounting table of chemistry.

To calculate reaction yields systematically, we set up a BCA table (Before, Change, After). Unlike math equations, a BCA table tracks the quantities of all substances at the same time:

Before (B): The starting mole quantities of reactants placed into the container before the reaction begins (products start at 0).
Change (C): The moles consumed (negative for reactants) or produced (positive for products). Crucially, the values in this row must scale in the exact ratio of the chemical equation coefficients.
After (A): The final mole quantities remaining when the reaction finishes (Before + Change).

Let's look at a balanced BCA table where 2.0 moles of CH₄ and 4.0 moles of O₂ are mixed. Because they 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)	—	—	—	—

2.0 mol CH₄ and 4.0 mol O₂ mixed — nothing has reacted yet, so products start at 0.

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)	—	—	—	—

The Change row is locked to the 1:2:1:2 coefficient ratio — reactants fall, products rise, in that exact proportion.

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

Before + Change = After. Because 2.0 mol CH₄ and 4.0 mol O₂ are in the exact 1:2 ratio, both reactants hit zero together — nothing is left over.

Notice that the Change row values (-2.0, -4.0, +2.0, +4.0) match the 1:2:1:2 coefficient ratio of the chemical equation. In chemistry, reactants are consumed, and products are built, in locked step.

03What runs out?

The limiting reactant sets the limit.

What happens if you mix 2.0 moles of CH₄ with 5.0 moles of O₂? You do not have enough methane to react with all of the oxygen. Methane will run 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 using a BCA table, you test which reactant would hit 0 first if fully consumed. The one that yields the smaller Change row multiplier is your limiting reactant. Once it hits 0 in the After row, the reaction stops, and the remaining excess reactant sits in the container unchanged.

04Interactive · BCA Table Simulator

Slide progress. Watch the atoms rearrange.

Adjust the starting moles of Methane (CH₄) and Oxygen (O₂) below. Slide the Reaction Progress slider to watch the molecules combust, and track the mole counts in the BCA table in real-time.

Methane Combustion: CH₄ + 2 O₂ → CO₂ + 2 H₂O

Initial Methane (CH₄) 2 mol

Initial Oxygen (O₂) 5 mol

Reaction Progress 0%

BCA Mole Accounting Table

Row	CH₄	+ 2 O₂
Before (B)	2.00	5.00
Change (C)	0.00	0.00
After (A)	2.00	5.00

Chemical Analysis

Ready to React

Slide the Reaction Progress bar to begin the reaction. Watch how reactant molecules in the chamber collide, tear apart, and assemble into CO₂ and H₂O.

05Predict · Reveal

Predict the reaction yield.

1. Limiting Reactant Setup: You mix 3.0 moles of CH₄ with 4.0 moles of O₂. According to the reaction recipe (CH₄ + 2 O₂ → CO₂ + 2 H₂O), which reactant is limiting and will run out first?

The Answer: Oxygen (O₂)

Each mole of methane requires 2 moles of oxygen. To react all 3.0 moles of CH₄, you would need 6.0 moles of O₂. Because you only have 4.0 moles of O₂, the oxygen will run out first. Methane is in excess (1.0 mole of CH₄ will remain unreacted).

2. Product Yield: In the same mixture (3.0 moles of CH₄ and 4.0 moles of O₂), what is the maximum amount of water (H₂O) that can be produced?

The Answer: 4.0 moles of H₂O

Because O₂ is limiting, the reaction stops when all 4.0 moles of O₂ are consumed. The coefficient ratio between O₂ and H₂O is 2:2 (or 1:1). Therefore, consuming 4.0 moles of O₂ produces exactly 4.0 moles of H₂O. (The excess methane does not contribute to further yield).

06Recall

Say it back.

Fill in the blanks to lock in the core terms. Matches are case-insensitive.

Stoichiometry Vocabulary

Fill in the terms.

1. The study of quantitative relationships in chemical reactions is: 2. The reactant that is completely consumed and limits the yield of product is the: 3. The reactant that is left over after a chemical reaction halts is the: 4. In a BCA table, the letter B stands for: 5. The values in the Change row of a BCA table must match the ratio of the equation's:

07Self-Explain

Why scale the Change row?

Why must the Change (C) row values in a BCA table scale in the exact ratio of the chemical equation coefficients, even if the starting Before (B) row does not?

Model Answer

The Before (B) row represents whatever random amounts of reactants are experimentally mixed together. However, molecules can only react on an individual, microscopic level according to the stoichiometry of the balanced chemical equation. Atoms rearrange in locked molecular proportions: for example, every 1 molecule of CH₄ that reacts requires exactly 2 molecules of O₂ to form 1 CO₂ and 2 H₂O. Because the reaction progress occurs strictly according to these recipe proportions, the Change (C) row—which represents the amounts actually reacting and forming—must scale in the exact ratio of the coefficients.

08Exam Practice

Solve a limiting reactant problem.

Give yourself a point for each idea you actually wrote down. The flag (⚑) marks the limiting reactant check.

Gen Chem · HS-PS1-7 · constructed response [4 marks]

A mixture contains 1.50 moles of methane (CH₄) and 2.00 moles of oxygen (O₂) gas. They react according to the equation: CH₄ + 2 O₂ → CO₂ + 2 H₂O.

Identify the limiting reactant, calculate the maximum moles of H₂O produced, and state the moles of excess reactant that will remain unreacted at the end of the reaction. Show all work (e.g., a sketch of a BCA table).

Mark Scheme — 4 marks

Sets up a BCA table or equivalent mathematical comparison showing starting values: Before: CH₄ = 1.50 mol, O₂ = 2.00 mol, H₂O = 0 mol.
Identifies O₂ as the limiting reactant because 1.50 mol CH₄ requires 3.00 mol O₂, but only 2.00 mol are available (or shows that O₂ runs out first under a Change multiplier of 1.00x). (⚑ Identifying the limiting reactant by comparing requirement to availability is required for this mark.)
Calculates the correct water yield: 2.00 moles of H₂O (derived from consuming 2.00 moles of O₂ in a 2:2 ratio).
Calculates the correct remaining excess reactant: 0.50 moles of CH₄ (Before 1.50 mol - Change 1.00 mol = After 0.50 mol).

Self-score: 4 = all four points · 3 = correct values but setup/limiting comparison is missing/unclear · 2 = identified limiting reactant but calculated incorrect products.

Why This Matters

Global crop fertilization relies on industrial stoichiometry. Synthesizing fertilizer ammonia via the Haber-Bosch process (N₂ + 3H₂ → 2NH₃) feeds half the world's population. Running the reaction in the exact 1:3 stoichiometric ratio prevents wasting valuable reactants, making global food production possible.