Balancing Equations and Solving Stoichiometric Problems

You now know that a balanced equation is the starting point for all stoichiometric calculations. But a key question remains: how do you actually produce a balanced equation in the first place? And once you have one, how do you put those coefficients to work on real numerical problems? This topic answers both questions, walking through the balancing method and then applying it across three fully worked problems.

How to Balance a Chemical Equation

The law of conservation of mass requires that the same number of atoms of each element appear on both sides of a chemical equation. Achieving this is called balancing the equation, and the most common approach is called the trial and error method (also known as balancing by inspection).

The process is straightforward: you place whole-number coefficients in front of chemical formulas and adjust them until every element’s atom count matches on both sides.

There is one rule you must never break: you cannot change the subscripts within a chemical formula to achieve balance. Subscripts define which substance you are dealing with. Altering a subscript changes the substance itself. For instance, turning $H_2O$ into $H_2O_2$ transforms water into an entirely different compound, hydrogen peroxide. Coefficients are the only thing you may adjust.

Balancing Simple Equations: Quick Examples

Some equations need only a single coefficient to reach balance. Seeing a few of these builds confidence before tackling a harder case.

Iron reacting with oxygen:

$4 \text{ Fe(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Fe}_2\text{O}_3\text{(s)}$

Atom check: 4 iron atoms on each side, 6 oxygen atoms on each side. Everything matches.

Magnesium reacting with oxygen:

$2 \text{ Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)}$

Atom check: 2 magnesium atoms on each side, 2 oxygen atoms on each side. Balanced.

Phosphorus reacting with oxygen:

Here is where a small adjustment is needed. Writing the unbalanced version first:

$\text{P}_4\text{(s)} + \text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}$

Phosphorus is fine (4 atoms on each side), but oxygen is not: there are only 2 on the left versus 10 on the right. Since each $\text{O}_2$ molecule supplies 2 oxygen atoms, you need 5 molecules to reach 10. Placing 5 in front of $\text{O}_2$:

$\text{P}_4\text{(s)} + 5\text{O}_2\text{(g)} \rightarrow \text{P}_4\text{O}_{10}\text{(s)}$

Oxygen now shows 10 atoms on each side. Balanced.

Balancing a More Complex Equation: Propane Combustion, Step by Step

When an equation involves more elements or larger coefficients, a systematic step-by-step approach is much more reliable than guessing. Here is the method, illustrated with the combustion of propane ($\text{C}_3\text{H}_8$).

Step 1: Write the correct formulas for all reactants and products

Propane burns in oxygen to produce carbon dioxide and water:

$\text{C}_3\text{H}_8\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$

This equation is unbalanced. The atom counts do not match on both sides.

Step 2: Balance carbon

There are 3 carbon atoms in propane on the left, but only 1 in a single $\text{CO}_2$ molecule on the right. Placing a coefficient of 3 in front of $\text{CO}_2$ fixes this:

$\text{C}_3\text{H}_8\text{(g)} + \text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$

Carbon is now balanced: 3 on each side.

Step 3: Balance hydrogen

Propane has 8 hydrogen atoms. Each water molecule contains 2 hydrogen atoms, so you need $8 \div 2 = 4$ molecules of water:

$\text{C}_3\text{H}_8\text{(g)} + \text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$

Hydrogen is now balanced: 8 on each side.

Step 4: Balance oxygen (saved for last)

Oxygen appears in two different products on the right side, which is exactly why it was left for last. Count up all the oxygen atoms on the right:

• From $3\text{CO}_2$: $3 \times 2 = 6$ oxygen atoms
• From $4\text{H}_2\text{O}$: $4 \times 1 = 4$ oxygen atoms
• Total on the right: $6 + 4 = 10$ oxygen atoms

On the left, each $\text{O}_2$ molecule provides 2 atoms. To reach 10, you need $10 \div 2 = 5$ molecules:

$\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$

Step 5: Verify the final equation

Always do a final check by counting every element on both sides:

Element	Left side	Right side
C	3	$3 \times 1 = 3$
H	8	$4 \times 2 = 8$
O	$5 \times 2 = 10$	$6 + 4 = 10$

All counts match. The equation is balanced.

A Practical Tip

Notice the strategy: carbon and hydrogen were each present in only one reactant and one product, so their coefficients could be locked down independently. Oxygen appeared in both $\text{CO}_2$ and $\text{H}_2\text{O}$ on the product side, making it the trickiest element. By handling it last, you avoid the frustration of balancing one element only to throw another off. This “save the multi-compound element for last” approach works well for most combustion and similar reactions.

Applying Stoichiometry: Solved Problems

With balancing under your belt, it is time to use balanced equations for actual calculations. The three problems below demonstrate how mole ratios from the equation translate into real mass-and-mole answers.

Problem 1.3: How Much Water Does Burning Methane Produce?

Question: Calculate the mass of water (in grams) produced by the combustion of 16 g of methane.

The balanced equation:

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$

Step 1:Convert the given mass to moles.

The molar mass of $\text{CH}_4$ is:

$12 + (4 \times 1) = 16 \text{ g/mol}$

So 16 g of methane is exactly:

$\frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol of } \text{CH}_4$

Step 2:Use the mole ratio from the equation.

The coefficients tell you that 1 mol of $\text{CH}_4$ produces 2 mol of $\text{H}_2\text{O}$.

$1 \text{ mol } \text{CH}_4 \rightarrow 2 \text{ mol } \text{H}_2\text{O}$

Step 3:Convert moles of product back to grams.

The molar mass of $\text{H}_2\text{O}$ is:

$(2 \times 1) + 16 = 18 \text{ g/mol}$

$2 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g}$

Answer: Burning 16 g of methane produces 36 g of water.

Problem 1.4: How Many Moles of Methane for a Given Mass of $\text{CO}_2$?

Question: How many moles of methane are required to produce 22 g of $\text{CO}_2$ after combustion?

The same balanced equation applies:

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$

Step 1:Convert the given mass of $\text{CO}_2$ to moles.

The molar mass of $\text{CO}_2$ is:

$12 + (2 \times 16) = 44 \text{ g/mol}$

$\text{Moles of } \text{CO}_2 = \frac{22 \text{ g}}{44 \text{ g/mol}} = 0.5 \text{ mol}$

Step 2:Apply the mole ratio.

The equation shows a 1:1 ratio between $\text{CH}_4$ and $\text{CO}_2$ (both have a coefficient of 1). So producing 0.5 mol of $\text{CO}_2$ requires exactly 0.5 mol of $\text{CH}_4$.

Answer: 0.5 mol of methane is required to produce 22 g of $\text{CO}_2$.

Problem 1.5: Finding the Limiting Reagent in Ammonia Synthesis

Question: 50.0 kg of $\text{N}_2$ and 10.0 kg of $\text{H}_2$ are mixed to produce $\text{NH}_3$. Calculate the amount of $\text{NH}_3$ formed and identify the limiting reagent.

This is the first problem where the two reactants are not in the perfect stoichiometric ratio, so one will run out before the other.

The balanced equation:

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$

This tells you that 1 mol of $\text{N}_2$ needs 3 mol of $\text{H}_2$ to produce 2 mol of $\text{NH}_3$.

Step 1:Convert both reactant masses to moles.

For nitrogen:

$\text{Moles of } \text{N}_2 = \frac{50.0 \text{ kg} \times 1000 \text{ g/kg}}{28.0 \text{ g/mol}} = \frac{50{,}000}{28.0} = 17.86 \times 10^2 \text{ mol}$

For hydrogen:

$\text{Moles of } \text{H}_2 = \frac{10.0 \text{ kg} \times 1000 \text{ g/kg}}{2.016 \text{ g/mol}} = \frac{10{,}000}{2.016} = 4.96 \times 10^3 \text{ mol}$

Step 2:Figure out which reactant runs out first.

Ask the question: “If all the $\text{N}_2$ were to react, how much $\text{H}_2$ would be needed?”

Using the 1:3 mole ratio:

$17.86 \times 10^2 \text{ mol } \text{N}_2 \times \frac{3 \text{ mol } \text{H}_2}{1 \text{ mol } \text{N}_2} = 5.36 \times 10^3 \text{ mol } \text{H}_2$

But the reaction only has $4.96 \times 10^3$ mol of $\text{H}_2$ available. That is less than the $5.36 \times 10^3$ mol required, so the hydrogen will run out before all the nitrogen is consumed.

$\text{H}_2$ is the limiting reagent. It determines how much product can form.

Step 3:Calculate the product using the limiting reagent.

Since the equation says 3 mol of $\text{H}_2$ produces 2 mol of $\text{NH}_3$:

$4.96 \times 10^3 \text{ mol } \text{H}_2 \times \frac{2 \text{ mol } \text{NH}_3}{3 \text{ mol } \text{H}_2} = 3.30 \times 10^3 \text{ mol } \text{NH}_3$

Step 4:Convert moles of product to mass.

The molar mass of $\text{NH}_3$ is:

$14.0 + (3 \times 1.0) = 17.0 \text{ g/mol}$

$3.30 \times 10^3 \text{ mol} \times 17.0 \text{ g/mol} = 56.1 \times 10^3 \text{ g} = 56.1 \text{ kg}$

Answer: The reaction produces 56.1 kg of $\text{NH}_3$, and $\text{H}_2$ is the limiting reagent.

Notice a key point: even though there was far less hydrogen by mass (10 kg vs. 50 kg of nitrogen), the decision about which reagent is limiting must always be made by comparing moles, not raw masses. Hydrogen molecules are extremely light (molar mass just 2.016 g/mol), so 10 kg of hydrogen actually provides a large number of moles. Even so, it was not quite enough to fully react with all the nitrogen.

Reactions in Solutions: A Preview

So far, all the reactions discussed have involved pure substances reacting in the solid or gaseous state. But the majority of reactions in a real laboratory take place in solutions, where substances are dissolved in a solvent.

When a substance is in solution, you need a way to express how much of it is present in a given amount of the solution. There are four standard methods:

1. Mass per cent (also called weight per cent, w/w %)
2. Mole fraction
3. Molarity
4. Molality

Each of these concentration measures captures the same basic idea, “how much substance is dissolved,” but uses a different basis for the measurement. These methods are explored in detail in the topics that follow.