Stoichiometry & Limiting Reagent

A balanced chemical equation gives quantitative information about the reaction. The coefficients tell you the mole ratios of all substances involved.

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$

This equation can be interpreted in three ways:

Mass check: 16 + 64 = 80 g (reactants) = 44 + 36 = 80 g (products) ✓ (Conservation of mass)

The key principle: always convert to moles first, use the mole ratio from the balanced equation, then convert back to the required unit.

Stoichiometry comes from two Greek roots: stoicheion (element) + metron (measure). It is the branch of chemistry that handles the quantitative relationships between substances in a chemical reaction — how much reactant is needed, how much product is formed, and in what proportions.

Starting from a balanced equation, every stoichiometric problem fits one of four standard types, classified by what is given and what is asked:

The strategy is the same in every type: convert everything to moles first, apply the mole ratios from the balanced equation, and convert back to whatever the question asks for. Once you internalise this, every stoichiometry problem reduces to the same three-step routine.

Consider the decomposition of potassium chlorate:

$\ce{2KClO3 -> 2KCl + 3O2}$

The coefficients give the mole ratios directly: 2 mol $\ce{KClO3}$ → 2 mol $\ce{KCl}$ + 3 mol $\ce{O2}$.

Converting to masses (using $M_{\ce{KClO3}} = 122.5$, $M_{\ce{KCl}} = 74.5$, $M_{\ce{O2}} = 32$):

$2 \times 122.5\text{ g } \ce{KClO3} \;\longrightarrow\; 2 \times 74.5\text{ g } \ce{KCl} + 3 \times 32\text{ g } \ce{O2}$

From this, the mass ratios are fixed by stoichiometry:

$\frac{\text{Mass of }\ce{KClO3}}{\text{Mass of }\ce{KCl}} = \frac{2 \times 122.5}{2 \times 74.5}$

Rearrange both sides — divide each by the corresponding molar mass — and the same equation reads, in moles:

$\frac{\text{Moles of }\ce{KClO3}}{2} = \frac{\text{Moles of }\ce{KCl}}{2}$

The coefficients of the balanced equation are the only numbers that matter. All mass-to-mass problems reduce to this kind of proportion.

When a reactant or product is a gas at known temperature and pressure, it is often easier to measure its volume than its mass. At STP, 1 mole of any gas occupies 22.4 L.

For the same decomposition:

$\ce{2KClO3 -> 2KCl + 3O2}$

The mass-to-volume ratios are fixed:

$\frac{\text{Mass of }\ce{KClO3}}{\text{Volume of }\ce{O2}\text{ at STP}} = \frac{2 \times 122.5}{3 \times 22.4}$

$\frac{\text{Mass of }\ce{KCl}}{\text{Volume of }\ce{O2}\text{ at STP}} = \frac{2 \times 74.5}{3 \times 22.4}$

These two relations let you find any one of mass of $\ce{KClO3}$, mass of $\ce{KCl}$, or volume of $\ce{O2}$ at STP — given any other. The 367.5 g $\ce{KClO3}$ worked example below uses exactly this idea.

Strip masses and volumes away, and what remains is the cleanest form of stoichiometry — the mole–mole relation. For the same reaction:

$\frac{\text{Moles of }\ce{KClO3}}{2} = \frac{\text{Moles of }\ce{KCl}}{2} = \frac{\text{Moles of }\ce{O2}}{3}$

This is just the balanced equation, read out loud.

General rule. For any balanced equation:

$aA + bB \longrightarrow cC + dD$

the following is always true:

$\frac{\text{Moles of A reacted}}{a} = \frac{\text{Moles of B reacted}}{b} = \frac{\text{Moles of C produced}}{c} = \frac{\text{Moles of D produced}}{d}$

where $a$, $b$, $c$, $d$ are the stoichiometric coefficients. This single relation handles every stoichiometric calculation. Convert everything to moles, apply the formula, and convert back at the end. Mole–mole is the foundation; mass–mass and mass–volume are just translations of it.

When reactants are not present in stoichiometric proportions, one of them will be completely consumed before the others. This is the limiting reagent (or limiting reactant):

• Limiting reagent — the reactant that is completely consumed; it limits the amount of product that can form
• Excess reagent — the reactant present in more than the stoichiometric amount; some remains unreacted after the reaction is complete

How to identify the limiting reagent:

1. Convert all given masses to moles
2. Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation
3. The reactant with the smallest quotient is the limiting reagent

Then:

• Use the moles of limiting reagent to calculate the amount of product
• Calculate the amount of excess reagent remaining

Two systematic methods work for any reaction. Both give the same answer; they suit different problem shapes.

Method 1 — Compare the given amount with the amount required. For $\ce{H2 + Cl2 -> 2HCl}$, the equation tells us 1 mol $\ce{H2}$ requires 1 mol $\ce{Cl2}$ (equivalent to 2 g : 71 g, or 1 vol : 1 vol). Whatever the input, ask: does one reactant fall short of what the other needs?

Method 2 — Divide moles by stoichiometric coefficient. The smaller quotient is the LR. For $\ce{N2 + 3H2 -> 2NH3}$:

Method 2 is fully systematic, doesn't depend on visual inspection, and works just as well for reactions with three or more reactants. Make Method 2 your default; use Method 1 when the required ratio is obvious by inspection.

In every stoichiometric calculation so far, we've assumed the reaction goes to completion — that every molecule of the limiting reagent is converted into product. Real reactions almost never behave that way. The amount of product collected in the lab is almost always less than the calculation predicts.

The amount actually collected is the actual yield. The maximum possible product, calculated from stoichiometry assuming perfect conversion, is the theoretical yield. The ratio between them — expressed as a percentage — is the percentage yield:

$\% \text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

Why actual yield falls short:

• The reaction doesn't go to completion. Some reactant remains unreacted at equilibrium.
• Material sticks to the glassware. Small amounts of product cling to the walls of flasks, condensers, and filter papers.
• Volatile products evaporate before being collected.
• Competing side reactions. Reactants take alternate routes, forming unintended by-products instead of the desired one.

For an industrial chemist, the entire job of process design is to maximise yield by minimising every one of these losses. A reaction with 40 % yield is often unprofitable; one with 95 % yield can scale to factories.