chemical equation with equal numbers of atoms for each element in the reactant and product

symbolic representation of a chemical reaction

number placed in front of symbols or formulas in a chemical equation to indicate their relative amount

complete ionic equation

chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions

chemical equation in which all reactants and products are represented as neutral substances

chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)

substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation

substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation

ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

← Back to Chemistry 2e

Chemistry 2e · 4.1 Writing and Balancing Chemical Equations

Balancing chemical equations: practice problems

6 problems from the book, with its answers. Check your working, or have any of them explained to you on the whiteboard.

How to approach these problems

Write every formula correctly first, then never touch the subscripts again: you balance only by changing the coefficients in front. Start with the element that appears in the fewest formulas, leave hydrogen and oxygen (and any element standing alone) for last, and treat a polyatomic ion that survives the reaction, such as NO₃⁻, as one unit. Count every atom on both sides after each change. If you end with a fraction, multiply everything through to get whole numbers, and check the coefficients can't all be divided by a common factor. The usual mistakes: changing a subscript to make the counts work, forgetting that O₂ and H₂ are diatomic, and not multiplying a coefficient across a parenthesis.

Exercises 1
What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?
Sign up free to have it checked
Exercises 3
Balance the following equations:
(a) $PCl_{5} (s) + H_{2} O (l) ⟶ POCl_{3} (l) + HCl (a q)$
Sign up free to have it checked
Exercises 5
Sign up free to have it checked
Exercises 7
Sign up free to have it checked
Exercises 9
Sign up free to have it checked
Exercises 11
Sign up free to have it checked

Key terms · 9

balanced equation: chemical equation with equal numbers of atoms for each element in the reactant and product
chemical equation: symbolic representation of a chemical reaction
coefficient: number placed in front of symbols or formulas in a chemical equation to indicate their relative amount
complete ionic equation: chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions
molecular equation: chemical equation in which all reactants and products are represented as neutral substances
net ionic equation: chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)
product: substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation
reactant: substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation
spectator ion: ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

Learn 4.1 on the whiteboard

The tutor teaches the whole section, drawing it out step by step and answering your questions as you go.

Read 4.1 in the textbook · This section on openstax.org

Balancing chemical equations: practice problems

6 problems from the book, with its answers. Check your working, or have any of them explained to you on the whiteboard.

How to approach these problems

Key terms · 9

balanced equation: chemical equation with equal numbers of atoms for each element in the reactant and product
chemical equation: symbolic representation of a chemical reaction
coefficient: number placed in front of symbols or formulas in a chemical equation to indicate their relative amount
complete ionic equation: chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions
molecular equation: chemical equation in which all reactants and products are represented as neutral substances
net ionic equation: chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)
product: substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation
reactant: substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation
spectator ion: ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

Learn 4.1 on the whiteboard

The tutor teaches the whole section, drawing it out step by step and answering your questions as you go.

Read 4.1 in the textbook · This section on openstax.org

Exercise 3, worked through on the whiteboard

A real lesson: the tutor solves it step by step on the board, the way it would with you.

Read what the tutor says

Hey Sam, I'm Theo, and we're going to break down these balancing problems step by step.

The big idea in chemical balancing is conservation of mass: every single atom that enters a reaction must come out the other side.

Looking at part a, we start by taking inventory of each element across the arrow.

Phosphorus and oxygen are already equal with one each, but hydrogen and chlorine both fall short on the product side.

Placing a coefficient of two in front of hydrochloric acid instantly delivers two hydrogens and raises the total chlorine count to five, balancing the whole reaction.

Now let's move directly to part b, which introduces a classic redox reaction featuring nitric acid.

Part b is easily the trickiest equation in the set because nitrogen and oxygen split into multiple products.

When pure inspection feels like chasing your tail, setting up a system of linear equations is the cleanest path forward.

Matching hydrogens shows that coefficient b equals two times d, and tracking copper gives a equals c.

Assigning d as four sets b to eight, which leads through the nitrogen and oxygen expressions directly to three for c, two for e, and three for a.

Next, we examine parts c, d, and e to look at the powerful odd-even oxygen strategy.

Let's take on parts c, d, and e, which showcase how diatomic elements dictate our choices.

Part c is straightforward: two hydrogen atoms and two iodine atoms enter, while hydrogen iodide contains only one of each.

Doubling hydrogen iodide immediately levels both elements without touching anything else.

Part d brings up the classic odd-even dilemma with iron and molecular oxygen.

Because oxygen gas only travels in pairs, it can never directly produce an odd total like three oxygen atoms.

Doubling iron three oxide forces the oxygen count up to six, which three diatomic oxygen molecules supply with ease.

That leaves four iron atoms to match on the left, finishing part d cleanly.

Part e uses that identical odd-even logic with hydrogen.

Water always supplies an even number of hydrogens, whereas the product side starts with three: one in sodium hydroxide and two in hydrogen gas.

Doubling sodium hydroxide brings the right-hand hydrogen total to four and demands two water molecules plus two sodium atoms on the left.

Up next, we tackle parts f, g, and h to handle ammonium dichromate decomposition and polyatomic counting.

Part f looks intimidating because of the complex formula, but expanding the parentheses reveals the structure clearly.

The subscript two outside the ammonium ion doubles every atom inside, giving two nitrogens and eight hydrogens on the reactant side.

Notice that both chromium and nitrogen already balance with coefficients of one, while the eight hydrogen atoms demand four water molecules.

Those four waters also bring four oxygens, which combine with the three oxygens in chromium three oxide to hit our target of seven perfectly.

Now we look at parts g and h, which both feature pure elemental forms.

In part g, tetraphosphorus supplies four phosphorus atoms, forcing a coefficient of four on phosphorus trichloride.

Four molecules of phosphorus trichloride contain twelve chlorine atoms in total, so six diatomic chlorine molecules lock in the balance.

Finally, platinum four chloride simply breaks down into one platinum atom and four chlorines, meaning two chlorine gas molecules complete the set.

Next, let's assemble our full master checklist and review the core heuristics we used across all eight reactions.

Here is the complete lineup of all eight balanced reactions, verified atom by atom against the textbook.

Looking back across the entire exercise, four essential heuristics make any balancing problem manageable.

First, always tackle elements that appear in only one species per side before touching anything else.

Second, leave pure elemental substances like oxygen or chlorine gas until the very end, because they let you tweak a single count without knocking other elements out of balance.

Third, whenever an odd total clashes with a diatomic molecule, doubling the compound containing the odd element clears the fraction immediately.

And finally, when products branch out in complicated reactions like copper with nitric acid, setting up linear variables eliminates all the guesswork.

Whenever you would like to test these out on a fresh practice equation or dig into the redox electron-transfer side of reaction b, just let me know.