Chemical Reactions and Stoichiometry

Definition: A chemical reaction is a process where reactants transform into new substances called products through the breaking and forming of chemical bonds; stoichiometry is the calculation of the quantitative relationships between reactants and products using a balanced equation.

How It Works

A chemical equation is only meaningful once it’s balanced.
The same number of atoms of each element must appear on both sides.
This reflects the law of conservation of mass, matter isn’t created or destroyed, only rearranged.
Balancing is done by adjusting coefficients, never subscripts, which would change what the substance actually is.

Reactions are commonly grouped by pattern:

Synthesis:          A + B → AB
Decomposition:       AB → A + B
Single displacement:  A + BC → AC + B
Double displacement:  AB + CD → AD + CB
Combustion:          CxHy + O2 → CO2 + H2O

Synthesis: two or more substances combine into one.
Decomposition: one substance breaks into two or more.
Single displacement: one element replaces another in a compound.
Double displacement: two compounds exchange partners, often driven by formation of a precipitate, gas, or water.
Combustion: a hydrocarbon plus O2 yields CO2 and H2O, releasing energy.
Once balanced, the coefficients give mole ratios, the exact proportion in which substances react or form.
Stoichiometry converts between mass, moles, and particles using molar mass (g/mol) and Avogadro’s number (6.022 × 10²³/mol) as conversion factors.
The balanced equation’s coefficients are the bridge between different substances.

Under the Hood

Worked example: limiting reagent and yield.

Given: ammonia synthesis, N2(g) + 3H2(g) → 2NH3(g), 28.0 g N2 (molar mass 28.0 g/mol) reacts with 6.00 g H2 (molar mass 2.02 g/mol)
Step 1: convert to moles

n(N2) = 28.0 g / 28.0 g/mol = 1.00 mol
n(H2) = 6.00 g / 2.02 g/mol = 2.97 mol

Step 2: find limiting reagent by comparing mole ratio needed (1:3) to what’s available

1.00 mol N2 needs 3.00 mol H2
only 2.97 mol H2 is available

Answer: H2 is the limiting reagent, just barely
Step 3: calculate theoretical yield of NH3 from the limiting reagent

n(NH3) = 2.97 mol H2 × (2 mol NH3 / 3 mol H2) = 1.98 mol
mass(NH3) = 1.98 mol × 17.0 g/mol = 33.7 g

Step 4: percent yield, given actual lab yield of 30.1 g

% yield = (actual / theoretical) × 100 = (30.1 / 33.7) × 100

Answer: 89.3% yield

Worked example: balancing a combustion reaction.

Given: propane combustion, C3H8 + O2 → CO2 + H2O
Step 1: balance carbon, 3 CO2 matches 3 C in propane
Step 2: balance hydrogen, 4 H2O matches 8 H in propane
Step 3: balance oxygen, right side now has 3(2) + 4(1) = 10 O atoms, so 5 O2 on the left
Answer:

C3H8 + 5O2 → 3CO2 + 4H2O

Why It Matters

Stoichiometry lets chemists and engineers predict exactly how much product a reaction will yield.
It predicts how much reactant to order, essential for lab-scale synthesis and industrial manufacturing.
Reagent cost and waste matter directly at industrial scale.
Pharmaceutical dosing reduces to the same mole-ratio logic.
Rocket propellant formulation reduces to the same mole-ratio logic.
Balancing a combustion engine’s air-fuel ratio reduces to the same mole-ratio logic.

Common Pitfalls

Adjusting subscripts instead of coefficients to balance an equation.
Changing a subscript changes the identity of the compound, H2O becomes H2O2, a different substance, rather than the amount of it.
Assuming the reactant present in the largest mass is automatically the limiting reagent.
It depends on the mole ratio required, not raw mass, as the ammonia example above shows.
Confusing theoretical yield, what stoichiometry predicts, with actual yield, what’s measured in the lab.
Percent yield is always actual/theoretical, capturing losses from side reactions, incomplete reactions, or purification.
Forgetting to convert to moles before applying mole ratios; using mass ratios directly gives wrong answers.
Losing track of significant figures through a multi-step calculation.
Rounding intermediate steps too early compounds error by the final answer.
Forgetting excess reagent remains in the product mixture.
A full analysis should account for how much of the non-limiting reagent is left over, not just how much product forms.

Comparison

Reaction Type	General Form	Example
Synthesis	A + B → AB	2H2 + O2 → 2H2O
Decomposition	AB → A + B	2H2O2 → 2H2O + O2
Single displacement	A + BC → AC + B	Zn + 2HCl → ZnCl2 + H2
Double displacement	AB + CD → AD + CB	AgNO3 + NaCl → AgCl + NaNO3
Combustion	CxHy + O2 → CO2 + H2O	CH4 + 2O2 → CO2 + 2H2O

Term	Meaning
Limiting reagent	Reactant that runs out first, caps the product amount
Excess reagent	Reactant with leftover amount after reaction stops
Theoretical yield	Maximum product predicted by stoichiometry
Actual yield	Product actually measured/isolated in the lab
Percent yield	Actual yield divided by theoretical yield, times 100

Real-World Application

Rocket propellant formulation is stoichiometry with zero room for error.

Many liquid-fueled rockets burn hydrazine (N2H4) with dinitrogen tetroxide (N2O4):

2N2H4 + N2O4 → 3N2 + 4H2O

Engineers calculate the precise mass ratio of fuel to oxidizer needed so both are consumed as close to simultaneously as possible.
Carrying excess of either wastes mass the rocket has to lift, directly cutting into payload capacity.
A shortfall of oxidizer leaves fuel unburned and reduces thrust.
Molar masses (N2H4 = 32.05 g/mol, N2O4 = 92.02 g/mol) combined with the 2:1 mole ratio give the exact mass ratio propellant tanks are sized around.

Example

Industrial ammonia synthesis, the Haber process, is stoichiometrically simple, N2 + 3H2 → 2NH3, but at plant scale, getting the 1:3 mole ratio of feedstock gases precisely right, and accounting for incomplete conversion per pass, determines how efficiently a multi-million-dollar reactor uses its natural gas feedstock, the usual hydrogen source.

FAQ

Why can’t you balance an equation by changing subscripts?

Subscripts define the compound’s actual formula and identity.
Changing them turns the substance into something chemically different rather than adjusting how much of the original substance is present.

Does a reaction with 100% theoretical atom economy always give 100% actual yield?

No. Atom economy is a theoretical measure of how much reactant mass ends up in the desired product versus by-products.
Actual yield also depends on real-world losses like incomplete reaction, side reactions, and purification, which atom economy doesn’t capture.

How do you find the limiting reagent with more than two reactants?

Calculate the amount of product each reactant could form individually, assuming it’s the only constraint.
The reactant that produces the least product is the limiting one.
Every other reactant has more than enough to keep pace with it.

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