Difficulty: Introductory | Prerequisites: mole concept, balancing chemical equations, molar mass calculations.
Stoichiometry is the quantitative backbone of chemistry. Every time you need to predict how much product a reaction makes, or figure out which reactant runs out first, you are doing stoichiometry. This topic covers mole-to-mole conversions using balanced equations and the limiting-reagent concept, both of which appear on virtually every general chemistry exam from here onward.
A balanced chemical equation tells you the exact mole ratios of reactants and products. When two reactants are mixed, one usually runs out first (the limiting reagent), and that one determines how much product you can make. Every stoichiometry problem boils down to converting what you are given into moles, using the ratio from the equation, and converting back to whatever the question asks for.
Stoichiometry
The area of chemistry that deals with the quantitative relationships between reactants and products in a balanced chemical equation. In simple terms, it is the maths of chemistry: how much of A do I need to make B, and how much B do I get?
Balanced chemical equation
An equation in which the number of atoms of each element is the same on both sides. The coefficients in front of each formula give the mole ratios. Think of it as a recipe. The coefficients are the recipe quantities, and you cannot change them without throwing the recipe off.
Mole ratio
The ratio of moles of one substance to moles of another, taken directly from the coefficients of the balanced equation. In simple terms, if the equation reads 3 NaOH + 1 H₃PO₄, the mole ratio of NaOH to H₃PO₄ is 3 : 1.
Limiting reagent (limiting reactant)
The reactant that is completely consumed first in a reaction, thereby determining the maximum amount of product that can form. Think of it as the ingredient you run out of first when cooking. If you have 2 slices of bread and 10 slices of cheese, bread is the limiting reagent for your sandwiches.
Excess reagent (excess reactant)
The reactant that is not fully consumed; some of it remains after the limiting reagent is used up. In simple terms, it is whatever is left over.
Theoretical yield
The maximum amount of product that can be formed from a given amount of limiting reagent, assuming perfect conversion and no losses. In simple terms, this is the best-case answer to "how much product can I make?"
Before you do any stoichiometry, the equation must be balanced. Unbalanced equations give wrong mole ratios.
Example: the combustion of magnesium is written as 2 Mg + O₂ → 2 MgO. Until you have those coefficients in place, you cannot use the equation for calculations.
Read the coefficients as mole ratios. In 3 NaOH + H₃PO₄ → Na₃PO₄ + 3 H₂O, the ratio of NaOH to Na₃PO₄ is 3 : 1. For every 3 mol of NaOH consumed, 1 mol of Na₃PO₄ forms.
To convert moles of one substance to moles of another, multiply by the appropriate mole ratio from the balanced equation.
When a problem gives you amounts of two (or more) reactants, one will run out first. That is the limiting reagent.
Method: for each reactant, calculate how many moles of product it could produce if it were fully consumed. The reactant that produces less product is the limiting reagent.
Alternative method: divide each reactant's given moles by its coefficient in the balanced equation. The reactant with the smallest result is the limiting reagent.
Example: 2.4 mol Mg and 10.0 mol O₂, equation 2 Mg + O₂ → 2 MgO. Divide: Mg gives 2.4/2 = 1.2, O₂ gives 10.0/1 = 10.0. Mg has the smaller value, so Mg is the limiting reagent.
Once you know the limiting reagent, use its moles and the mole ratio to find the moles of product.
Example continued: 2.4 mol Mg × (2 mol MgO / 2 mol Mg) = 2.4 mol MgO.
The excess reagent's leftover can also be calculated: O₂ needed = 2.4 mol Mg × (1 mol O₂ / 2 mol Mg) = 1.2 mol O₂ used. Remaining O₂ = 10.0 − 1.2 = 8.8 mol.
Given: 35.60 mol NaOH and 10.80 mol H₃PO₄. Equation: 3 NaOH + H₃PO₄ → Na₃PO₄ + 3 H₂O.
Divide by coefficients: NaOH gives 35.60/3 = 11.87, H₃PO₄ gives 10.80/1 = 10.80. H₃PO₄ is smaller, so H₃PO₄ is the limiting reagent.
Moles of Na₃PO₄ = 10.80 mol H₃PO₄ × (1 mol Na₃PO₄ / 1 mol H₃PO₄) = 10.80 mol Na₃PO₄.
Mole ratio (from balanced equation)
For aA + bB → cC + dD, the ratio of A to C is a : c. To convert moles of A to moles of C: moles of C = moles of A × (c / a).
Grams to moles
n = mass (g) ÷ molar mass (g/mol)
Moles to grams
mass (g) = n × molar mass (g/mol)
Limiting reagent test
For each reactant, compute: (moles available) ÷ (its coefficient). The smallest result identifies the limiting reagent.
Theoretical yield of product
moles of product = moles of limiting reagent × (product coefficient / limiting reagent coefficient)
Limiting-reagent analysis is how chemical engineers decide how much raw material to order. If Reactant A is expensive and Reactant B is cheap, they will deliberately use excess B to ensure all of A is consumed. Pharmaceutical manufacturing relies on this to maximise yield of costly drug intermediates.
Stoichiometry also underpins everyday chemistry like adjusting baking recipes (double the recipe means double every ingredient) and calculating fuel requirements for rockets, where the ratio of fuel to oxidiser must be precise.
Students often assume the reactant present in the smaller number of moles is automatically the limiting reagent. This is wrong. You must account for the stoichiometric coefficients. A reactant present in fewer moles can still be in excess if its coefficient is small.
Another common error is forgetting to balance the equation before computing mole ratios. An unbalanced equation gives incorrect ratios and every subsequent calculation will be wrong.
Some students confuse "limiting reagent" with "the reagent that limits the reaction rate." In stoichiometry, the limiting reagent is purely about quantity, not speed.
When asked "how many moles of product form," students sometimes report the moles of excess reagent consumed instead of the product. Read the question twice.
⚠️ Limiting-reagent problems appear on almost every general chemistry exam. The divide-by-coefficient method is the fastest approach.
⚠️ Always check that your equation is balanced before doing anything else. An unbalanced equation makes every number wrong.
⚠️ Some problems ask for the amount of excess reagent remaining. To get this, calculate how much of the excess reagent reacted (using the limiting reagent's moles and the mole ratio), then subtract from the starting amount.
⚠️ Watch your units. If the problem gives grams, convert to moles before using the mole ratio, and convert back to grams if the answer asks for grams.
True or false: the reactant with the fewest moles is always the limiting reagent.
Answer: False. You must divide each reactant's moles by its coefficient before comparing.
Fill in the blank: in the equation 2 Mg + O₂ → 2 MgO, the mole ratio of Mg to MgO is ________.
Answer: 2 : 2, which simplifies to 1 : 1.
True or false: if the equation is not balanced, your stoichiometric calculations will still be correct as long as you use molar masses carefully.
Answer: False. Mole ratios come from a balanced equation. Without it, every ratio is wrong.
Fill in the blank: once you identify the limiting reagent, you calculate the moles of product using the ________ and the mole ratio.
Answer: moles of the limiting reagent.
Q: Balance the equation ___ Mg + ___ O₂ → ___ MgO, then calculate the moles of MgO that can form from 2.4 mol Mg and 10.0 mol O₂.
A: Balanced equation: 2 Mg + O₂ → 2 MgO. Divide by coefficients: Mg gives 2.4/2 = 1.2, O₂ gives 10.0/1 = 10.0. Mg is the limiting reagent. Moles of MgO = 2.4 mol Mg × (2 mol MgO / 2 mol Mg) = 2.4 mol MgO.
Q: 35.60 mol of NaOH is reacted with 10.80 mol of H₃PO₄. How many moles of Na₃PO₄ are formed? (3 NaOH + H₃PO₄ → Na₃PO₄ + 3 H₂O)
A: Divide by coefficients: NaOH gives 35.60/3 = 11.87, H₃PO₄ gives 10.80/1 = 10.80. H₃PO₄ is the limiting reagent. Moles of Na₃PO₄ = 10.80 mol H₃PO₄ × (1 mol Na₃PO₄ / 1 mol H₃PO₄) = 10.80 mol Na₃PO₄.
Q: In the reaction above, how many moles of NaOH remain unreacted?
A: NaOH consumed = 10.80 mol H₃PO₄ × (3 mol NaOH / 1 mol H₃PO₄) = 32.40 mol. Remaining NaOH = 35.60 − 32.40 = 3.20 mol.
Q: If 4.8 mol of Mg reacted with 2.0 mol of O₂ (2 Mg + O₂ → 2 MgO), how many moles of MgO form and which reagent is in excess?
A: Divide by coefficients: Mg gives 4.8/2 = 2.4, O₂ gives 2.0/1 = 2.0. O₂ is the limiting reagent. Moles of MgO = 2.0 mol O₂ × (2 mol MgO / 1 mol O₂) = 4.0 mol MgO. Mg is in excess: Mg consumed = 2.0 × 2 = 4.0 mol, remaining = 4.8 − 4.0 = 0.8 mol.
Stoichiometry and limiting reagents feed directly into solution stoichiometry (molarity and dilution calculations) and gas stoichiometry (using PV = nRT alongside mole ratios). If you are solid on the mole-ratio method here, those extensions add one extra conversion step each.
This topic also connects back to the energy and enthalpy material. Combining ΔH with a limiting-reagent calculation lets you predict both how much product forms and how much heat is released or absorbed, which is exactly what industrial chemists need to design safe processes.
Percentage yield, which compares the actual yield from an experiment to the theoretical yield from stoichiometry, is the next step after this topic and appears in most lab reports.
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