"How to write and understand chemical equations"

Seven equations are presented, but approached in the following way

The individual symbols and formulae are explained

The word equation is presented to summarise the change of reactants to products.

A balanced 'picture' equation which helps you understand reading formulae and atom counting to balance the equation.

The fully written out symbol equation with state symbols (often optional for starter students).

For any reaction, what you start with are called the reactants, and what you form are called the products.

So any chemical equation shows in some way the overall chemical change of  ...

REACTANTS ==> PRODUCTS, which can be written in words or symbols/formulae.

It is most important you read about formula in an earlier section, but examples are explained on this page too.

In the equations outlined below several things have been deliberately simplified. This is to allow the 'starter' chemistry student to concentrate on understanding formulae and balancing chemical equations. Some teachers may disagree with this approach BUT my simplifications are:

The word 'molecule' is sometimes loosely used to mean a 'formula'.

The real 3D shape of the 'molecule' and the 'relative size' of the different element atoms is ignored.

If the compound is ionic, the ion structure and charge is ignored, its just treated as a 'formula'.

 

==> means the direction of change from reactants == to ==> products

No symbols or numbers are used in word equations.

Always try to fit all the words neatly lined up from left to right, especially if its a long word equation.

Writing the correct symbol or formula for each equation component.

Numbers in a formula are written as subscripts after the number of atoms of the element concerned

e.g. H₂SO₄ means 2 H's, 1 S and 4 O's

or the subscript number can double, treble etc. a part of the formula

e.g. Ca(OH)₂ means 1 Ca and 2 OHs' (or 2 Os' and 2 Hs' in total)

Numbers before a formula double or treble it etc.

e.g. 2NaCl means 2 Na's and 2 Cl's in total

or 2H₂SO₄ means 2 x H₂SO₄ =  4 H's, 2 S's and 8 O's in total of 2 H₂'s and 2 SO₄'s

Its quite handy to think in different ways to balance an equation, but ultimately, and the most logically, its all about counting atoms correctly and making sure you have the same number of atoms of each element on each side of the equations BUT this only works if all the chemical formulae are correct in the equation!

If the number is 1 itself, by convention, no number is shown in a formula or before a formula.

Using numbers if necessary to balance the equation, this is a matter of 'trial and error'.

If you seem deficient on one side of the equation in terms of a particular atom (element), you may need to increase a balancing number on the other side of the equation.

Also, you must use the smallest possible whole numbers (integers) to balance an equation, its best to avoid half-numbers if you can. dealing in half-molecules may seem a bit strange!

If all is correct, then the sum of atoms for each element should be the same on both side of the equation arrow .....

in other words: atoms of products = atoms of reactants

This is a chemical conservation law of atoms and later it may be described as the 'law of conservation of mass'.

the 7 equations are first presented in 'picture' style and then written out fully with state symbols

The individual formulas involved and the word equations will be been presented in the examples below.

NEVER alter a formula to balance an equation! BUT use the CORRECT FORMULA and only put NUMBERS BEFORE THE FORMULA if needed to balance the number of atoms to balance the equation.

You do need to be able to read and understand a formula and I've included some reminders and explanations in the examples of how to construct and balance the 7 chemical equations described and explained in detail below.

PRACTICE QUESTIONS - on words and symbol equations (on other pages)

Multiple choice quiz on balancing numbers

Balancing number/formula-fill exercises

Reactions of acids with metals, oxides, hydroxides, carbonates and ammonia.

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3.1d EXAMPLES of CONSTRUCTING WORD or SYMBOL EQUATIONS

Remember from the 'Law of Conservation of Mass' the mass of products = mass of original reactants, which means that the number of atoms of each element in the reactants must be equal to those in the products and that is the basis of writing a correctly balanced symbol equation, BUT don't forget, you must write the correct formula for each species in the equation, otherwise you may write a correctly balanced equation which is totally wrong! so beware!

Balancing equations example  3.1d (1) iron + sulfur ==>

A single symbol means an uncombined single atom of the element, or Fe 1 atom of iron, or S 1 atom of sulfur (2Fe would mean two atoms, 5S would mean five sulfur atoms etc.)

or the formula FeS means one atom of iron is chemically combined with 1 atom of sulfur to form the compound called iron sulfide

iron + sulfur ==> iron sulfide

on average one atom of iron chemically combines with one atom of iron forming one molecule of iron sulfide

two elements chemically combining to form a new compound

Fe + S ==> FeS

Fe(s) + S(s) ==> FeS(s)   (with state symbols)

Atom balancing, sum left = sum right: 1Fe + 1S = (1Fe combined with 1S)

There is no need for any balancing numbers in this equation

For a balanced equation on both sides of the equation you should have 1 iron atom and 1 sulfur atom combined in their particular way in the reactants or products

All the reactants (what you start with) and all the products (what is formed) are all solids in this case.

When first learning symbol equations you probably won't use state symbols like (s) at first (see end note).

Balancing equations example 3.1d (2) sodium hydroxide + hydrochloric acid ==>

or the formula NaOH means 1 atom of sodium is combined with 1 atom of oxygen and 1 atom of hydrogen to form the compound called sodium hydroxide

the reactants are one molecule of sodium hydroxide and one molecule of hydrochloric acid

the products are one molecule of sodium chloride and one molecule of water

all the chemical reactants and products involved are compounds

NaOH + HCl ===> NaCl + H₂O

NaOH(aq) + HCl(aq) ===> NaCl(aq) + H₂O(l)   (with state symbols)

atom balancing, sum left = right: (1Na + 1O + 1H) + (1H + 1Cl) = (1Na + 1Cl) + (2H + 1O)

For a balanced equation on both sides of the equation you should have 1 sodium atom, 1 oxygen atom, 1 chlorine atom and 2 hydrogen atoms combined in their particular way in the reactants or products.

Note the subscript 2 after the H in water means two atoms of that element.

There is no need for any balancing numbers in this equation

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Balancing equations example 3.1d (3) magnesium + hydrochloric acid ==>

Thinking sequence:

Mg + HCl ==> MgCl₂ + H₂ (short of H and Cl on left)

Mg + 2HCl ==> MgCl₂ + H₂ (doubling the HCl sorts this out)

or the symbol Mg means 1 atom of the element called magnesium

one atom of magnesium reacts with two molecules of hydrochloric acid

the products are one molecule of magnesium chloride and one molecule of hydrogen

Mg and H-H (H₂) are elements, H-Cl (HCl) and Cl-Mg-Cl (MgCl₂) are compounds

So the balanced equation is ...

Mg + 2HCl ==> MgCl₂ + H₂

Mg(s) + 2HCl(aq) ==> MgCl₂(aq) + H₂(g)   (with state symbols)

and you should realise why the 2 is needed

atom balancing, sum left = right: (1Mg) + 2 x (1H + 1Cl) = (1Mg + 2Cl) + (2H)

For a balanced equation on both sides of the equation you should have 1 magnesium atom, 2 hydrogen atoms and 2 chlorine atoms combined in their particular way in the reactants or products. You can only get the balance here by putting a 2 in front of the HCl formula because you need 2 Cl's to make the MgCl₂.

Balancing equations example 3.1d (4) copper carbonate + sulfuric acid ==>

or the formula CuCO₃ means one formula of the compound called copper carbonate, made up of one atom of copper is combined with one atom of carbon and three atoms of oxygen (subscript 3 after the O) to form the compound copper carbonate

  copper carbonate  + sulfuric acid ==> copper  sulfate + water + carbon dioxide

the reactants are one formula of copper carbonate and one molecule of sulfuric acid

the products are one formula of copper sulfate, one molecule of water and one molecule of carbon dioxide

all molecules are compounds in this reaction

No balancing numbers are needed, all the atoms add up on both sides, so the balanced equation is ...

CuCO₃ + H₂SO₄ ==> CuSO₄ + H₂O + CO₂

CuCO₃(s) + H₂SO₄(aq) ==> CuSO₄(aq) + H₂O(l) + CO₂(g)  (with state symbols)

balancing sum left = sum right: (1Cu + 1C + 3O) + (2H + 1S + 4O) = (1Cu + 1S + 4O) + (2H + 1O) + (1C + 2O)

For a balanced equation on both sides of the equation you should have 1 copper atom, 1 carbon atom, 7 oxygen atoms, 2 hydrogen atoms, 1 sulfur atom combined in their particular way in the reactants or products

There is no need for any balancing numbers in this equation

Balancing equations example 3.1d (5) burning a hydrocarbon e.g. methane + oxygen ==>

Thinking sequence:

CH₄ + O₂ ==> CO₂ + H₂O  (short of H on right, need to add H to right, short of O on left)

CH₄ + O₂ ==> CO₂ + 2H₂O  (doubling the water sorts out the Hs, but still short of O on left)

CH₄ + 2O₂ ==> CO₂ + 2H₂O  (then also doubling the O₂ on left sorts this out)

or the formula CH₄ means one molecule of the compound called methane which is made of one atom of carbon combined with four atoms of hydrogen. Oxygen is an element and the two products, carbon dioxide and water, are compounds.

one molecule of methane is completely burned by two molecules of oxygen

to form one molecule of carbon dioxide and two molecules of water

So the balanced equation is ...

CH₄ + 2O₂ ===> CO₂ + 2H₂O

CH₄(g) + 2O₂(g) ===> CO₂(g) + 2H₂O(l)   (with state symbols)

atom balancing, sum left = sum right: (1C + 4H) + 2 x (2O) = (1C + 2O) + 2 x (2H + 1O)

For a balanced equation on both sides of the equation you should have 1 carbon atom, 4 hydrogen atoms, 4 oxygen atoms combined in their particular way in the reactants or products.

You can only get this to balance by having a 2 in front of the O₂ and 2 in front of the CO₂, you need an O₂ to make CO₂ and another O₂ to convert the H₄ into 2H₂O

Note the 2 before the O₂ and H₂O doubles the number of these molecules to balance the equation.

Using the complete combustion of a hydrocarbon as an example, its worth mentioning the use of 'fractions' of a formula in balancing equations

e.g. ethane + oxygen ==> carbon dioxide + water

C₂H₆ + O₂ ==> CO₂ + H₂O (short of C and H on right, short of O on left)

C₂H₆ + O₂ ==> 2CO₂ + 3H₂O (adding the 2 and 3 on right nearly gets it balanced, but not for O)

You need 7 oxygens to balance the left, but they come in pairs, so you can say 3½ oxygen molecules

C₂H₆ + 3½O₂ ==> 2CO₂ + 3H₂O (and this sorts out a balanced equation)

At GCSE level, fractions of a molecule are usually avoided, but not at A level, when you should automatically think of any equation in terms of mole proportions of reactants and products.

The equation above, can be written without fractions, by simply numerically doubling everything up, giving

2C₂H₆ + 7O₂ ==> 4CO₂ + 6H₂O (and its all the same ratio in the end)

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Balancing equations example 3.1d (6) magnesium hydroxide + nitric acid ==>

Thinking:

Mg(OH)₂ + H₂SO₄ ==> MgSO₄ + H₂O (short of H and O on right)

Mg(OH)₂ + H₂SO₄ ==> MgSO₄ + 2H₂O (just doubling the water product sorts it out)

or the formula Mg(OH)₂ is the compound magnesium hydroxide made up of one magnesium, two oxygen and two hydrogen atoms BUT the OH is a particular combination called hydroxide  within a compound, so it is best to think of this compound as a combination of an Mg and two OH's, hence the use of the (brackets). The subscripted 2 doubles everything in the brackets.

Note that here the subscript 2 after the (NO₃) in the magnesium nitrate means everything in the brackets is doubled.

one formula of magnesium hydroxide reacts with two molecules of nitric acid to form one formula of magnesium nitrate and two molecules of water (all reactants and products are compounds)

Mg(OH)₂ + 2HNO₃ ==> Mg(NO₃)₂ + 2H₂O

Mg(OH)₂(aq) + 2HNO₃(aq) ==> Mg(NO₃)₂(aq) + 2H₂O(l)  (with state symbols)

atom balancing, sum left = sum right: (1Mg + 2O + 2H) + 2 x (1H + 1N + 3O) = (1Mg + 2N + 6O) + 2 x (2H + 1O)

For a balanced equation on both sides of the equation you should have 1 magnesium atom, 8 oxygen atoms, 4 hydrogen atoms, 2 nitrogen atoms combined in their particular way in the reactants or products

to balance this equation you need a 2 in front of the HNO₃ and a 2 in front of the H₂O, the 2s come from the 2 OH becoming 2 H₂Os

Balancing equations example 3.1d (7) aluminium oxide + sulfuric acid

Thinking:

Al₂O₃ + H₂SO₄ ==> Al₂(SO₄)₃ + H₂O (initially just think short of SO₄ on left)

Al₂O₃ + 3H₂SO₄ ==> Al₂(SO₄)₃ + H₂O (tripling the H₂SO₄ sorts this out, but short H and O on right)

Al₂O₃ + 3H₂SO₄ ==> Al₂(SO₄)₃ + 3H₂O (tripling the H₂O sorts this out for both H and O)

or the formula Al₂O₃ means one formula of the compound called aluminium oxide, made up of two atoms of aluminium Al and three atoms of oxygen O

one formula of aluminium oxide reacts with three molecules of sulfuric acid

to form one formula of aluminium sulfate and three molecules of water, all reactants and products are compounds

note the first use of numbers (3) for the sulfuric acid and water!

so picture three of them in your head, otherwise the picture gets a bit big!

Al₂O₃ + 3H₂SO₄ ==> Al₂(SO₄)₃ + 3H₂O

Al₂O₃(s) + 3H₂SO₄(aq) ==> Al₂(SO₄)₃(aq) + 3H₂O(l)   (with state symbols)

atom balancing, sum left = sum right: (2Al + 3O) + 3 x (2H + 1S + 4O) = (2Al + 3S + 12O) + 3 x (2H + 1O)

For a balanced equation on both sides of the equation you should have 2 aluminium atoms, 15 oxygen atoms, 6 hydrogen atoms, 3 sulfur atoms combined in their particular way in the reactants or products

This is quite an awkward equation to balance, a bit of real trial and error, but two 3s in the right place will do it.

The best clue here is that you need 3 x SO₄ for the aluminium sulfate, so you need 3 of the H₂SO₄

GCSE-AS-A2-IB note: Aluminium sulfate is actually an ionic compound (Al³⁺)₂(SO₄^2-)₃

Extra NOTE 1 Reversible Reactions

The sign means a reversible reaction, it can be made to go the 'other way' if the conditions are changed. Example:

nitrogen + hydrogen ammonia

N_2(g) + 3H_2(g) 2NH_3(g)   (with state symbols)

balancing: 2 nitrogen's and 6 hydrogen's on both sides of equation

Extra NOTE 2 State Symbols

The use state symbols X_(?) of reactants or products in equations, used to denote the physical state of reactants and products

(g) means gas, (l) means liquid, (s) means solid

and (aq) means aqueous solution, which means the substance is dissolved in water

e.g. carbon dioxide gas CO_2(g), liquid water H₂O_(l), solid sodium chloride 'salt' NaCl_(s)

and copper sulfate solution CuSO_4(aq)

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What is an 'ionic equation'? How do we construct and write ionic equations?

In many reactions only certain ions change their 'chemical state' but other ions remain in exactly the same original physical and chemical state.

The ions that do not change physically or chemically are called 'spectator ions'.

The ionic equation represents the 'actual' chemical change and omits ALL of the spectator ions.

Five types of examples of ionic equations are presented below including neutralisation, salt precipitation and redox equations.

(1) Acid-base neutralisation reactions:

Acids can be defined as proton donors. A base can be defined as a proton acceptor.

e.g. any acid-alkali neutralisation involves the hydroxide ion is (base) and this accepts a proton from an acid.

HCl_(aq) + NaOH_(aq) ===> NaCl_(aq) + H₂O_(l) which can be re-written ionically as

H⁺Cl^-_(aq) + Na⁺OH^-_(aq) ===> Na⁺Cl^-_(aq) + H₂O_(l)

or: H⁺_(aq) + Cl^-_(aq) + Na⁺_(aq) + OH^-_(aq) ===> Na⁺_(aq) + Cl^-_(aq) + H₂O_(l)

H⁺_(aq) + OH^-_(aq) ===> H₂O_(l) which is the ionic equation for neutralisation

the unchanged spectator ions are chloride Cl^- and sodium Na⁺ and can be omitted to leave the true ionic equation.

(2) Insoluble salt formation - salt precipitation reactions:

An insoluble salt is made by mixing two solutions of soluble compounds to form the insoluble compound in a process called 'precipitation'. A precipitation reaction is generally defined as 'the formation of an insoluble solid on mixing two solutions or a bubbling a gas into a solution'.

See preparation of insoluble salts

(a) Silver chloride is made by mixing solutions of solutions of silver nitrate and sodium chloride.

silver nitrate + sodium chloride ==> silver chloride + sodium nitrate

AgNO_3(aq) + NaCl_(aq) ==> AgCl_(s) + NaNO_3(aq)

in terms of ions it could be written as

Ag⁺NO₃^-_(aq) + Na⁺Cl^-_(aq) ==> AgCl_(s) + Na⁺NO₃^-_(aq)

or: Ag⁺_(aq) + NO₃^-_(aq) + Na⁺_(aq) + Cl^-_(aq) ==> AgCl_(s) + Na⁺_(aq) + NO₃^-_(aq)

but the omitted spectator ions are nitrate NO₃^- and sodium Na⁺ which do not change at all,

so the ionic equation is simply: Ag⁺_(aq) + Cl^-_(aq) ==> AgCl_(s)

Note that ionic equations omit ions that do not change there chemical or physical state.

In this case the nitrate (NO₃^-_(aq)) and sodium (Na⁺_(aq)) ions do not change physically or chemically and are called spectator ions,

BUT the aqueous silver ion, Ag⁺_(aq), combines with the aqueous chloride ion, Cl^-_(aq), to form the insoluble salt silver chloride, AgCl_(s), thereby changing their states both chemically and physically.

 

(b) Lead(II) iodide, a yellow precipitate (insoluble in water!) can be made by mixing lead(II) nitrate solution with e.g. potassium iodide solution.

lead(II) nitrate + potassium iodide ==> lead(II) iodide + potassium nitrate

Pb(NO₃)_2(aq) + 2KI_(aq) ==> PbI_2(s) + 2KNO_3(aq)

which can be written as

Pb²⁺_(aq) + 2NO₃^-_(aq) + 2K⁺_(aq) + 2I^-_(aq) ==> PbI_2(s) + 2K⁺_(aq) + 2NO₃^-_(aq)

the ionic equation is: Pb²⁺_(aq) + 2I^-_(aq) ==> PbI_2(s)

because the omitted unchanged spectator ions are nitrate NO₃^- and potassium K⁺.

 

(c) Calcium carbonate, a white precipitate, forms on e.g. mixing calcium chloride and sodium carbonate solutions ...

calcium chloride + sodium carbonate ===> calcium carbonate + sodium chloride

CaCl_2(aq) + Na₂CO_3(aq) ===> CaCO_3(s) + 2NaCl_(aq)

Ca²⁺_(aq) + 2Cl^-_(aq) + 2Na⁺_{(aq) +} CO₃^2-_(aq) ===> CaCO_3(s) + 2Na⁺_(aq) + 2Cl^-_(aq)

ionically: Ca²⁺_(aq) + CO₃^2-_(aq) ===> CaCO_3(s)

because the omitted unchanged spectator ions are chloride Cl^- and sodium Na⁺.

 

(d) Barium sulfate, a white precipitate, forms on mixing e.g. barium chloride and dilute sulfuric acid ...

barium chloride + sulfuric acid ==> barium sulfate + hydrochloric acid

BaCl_2(aq) + H₂SO_4(aq) ==> BaSO_4(s) + 2HCl_(aq)

Ba²⁺_(aq) + 2Cl^-_(aq) + 2H⁺_{(aq) +} SO₄^2-_(aq) ==> BaSO_4(s) + 2H⁺_(aq) + 2Cl^-_(aq)

ionic equation: Ba²⁺_(aq) + SO₄^2-_(aq) ==> BaSO_4(s)

because the unchanged spectator ions are chloride Cl^- and hydrogen H⁺.

(e) Calcium hydroxide is only slightly soluble in water, so adding any alkali to a soluble calcium salt solution results in the precipitation of calcium hydroxide (s) e.g.

calcium chloride  +  sodium hydroxide  ==>  calcium hydroxide  +  sodium chloride

CaCl₂(aq)  +  2NaOH(aq)  ==>  Ca(OH)₂(s)  +  2NaCl(aq)

In terms of all the ions present

Ca²⁺(aq)  +  2Cl^-(aq)  +  2Na⁺(aq)  +  2OH^-(aq)  ==>  Ca(OH)₂(s)  +  2Na⁺(aq)  +  2Cl^-(aq)

So the ionic equation becomes simply:

Ca²⁺(aq)  +   2OH^-(aq)  ==>  Ca(OH)₂(s)

because the unchanged spectator ions are the sodium ions and chloride ions.

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(3) Redox reaction analysis - reactions involving an oxidation and reduction:

(a) magnesium + iron(II) sulfate ==> magnesium sulfate + iron

Mg_(s) + FeSO_4(aq) ==> MgSO_4(aq) + Fe_(s)

this is the 'ordinary molecular' equation for a typical metal displacement reaction, but this does not really show what happens in terms of atoms, ions and electrons, so we use ionic equations like the one shown below.

Mg_(s) + Fe²⁺SO₄^2-_(aq) ==> Mg²⁺SO₄^2-_(aq) + Fe_(s)

The sulfate ion SO₄^2-_(aq) is the spectator ion, because it doesn't change in the reaction and can be omitted from the ionic equation. No electrons show up in the full equations because electrons lost by Mg must equal the electrons gained by Fe.

so the ionic-redox equation is

Mg_(s) + Fe²⁺_(aq) ==> Mg²⁺_(aq) + Fe_(s)

Mg oxidised by electron loss, Fe²⁺ reduced by electron gain

(b) zinc + hydrochloric acid ==> zinc chloride + hydrogen

Zn_(s) + 2HCl_(aq) ==> ZnCl_2(aq) + H_2(g)

Zn_(s) + 2H⁺_(aq) + 2Cl^-_(aq) ==> Zn²⁺_(aq) + 2Cl^-_(aq) + H_2(g)

the chloride ion Cl^- is the spectator ion

Zn_(s) + 2H⁺_(aq) ==> Zn²⁺_(aq) + H_2(g)

Zinc atoms, Zn, oxidised by electron loss and hydrogen ions, H⁺, are reduced by electron gain

(c) copper + silver nitrate ==> silver + copper(II) nitrate

Cu_(s) + 2AgNO_3(aq) ==> 2Ag + Cu(NO₃)_2(aq)

the nitrate ion NO₃^- is the spectator ion

Cu_(s) + 2Ag⁺_(aq) ==> 2Ag_(s) + Cu²⁺_(aq)

Cu oxidised by electron loss, Ag⁺ reduced by electron gain

(d) halogen (more reactive) + halide salt (of less reactive halogen) ==> halide salt (of more reactive halogen) + halogen (less reactive)

X_2(aq) + 2K⁺Y^-_(aq) ==> 2K⁺X^-_(aq) + Y_2(aq)

X_2(aq) + 2Y^-_(aq) ==> 2X^-_(aq) + Y_2(aq)

the potassium ion K⁺ is the spectator ion

halogen X is more reactive than halogen Y, F > Cl > Br > I

X is the oxidising agent (electron acceptor, so is reduced)

KY or Y^- is the reducing agent (electron donor, so is oxidised)

(4) Ion Exchange Resins:

Ion exchange polymer resin columns hold hydrogen ions or sodium ions. These can be replaced by calcium and magnesium ions when hard water passes down the column. The calcium or magnesium ions are held on the negatively charged resin. The freed hydrogen or sodium ions do not form a scum with soap.

e.g. 2[resin]^-H⁺_(s) + Ca²⁺_(aq) ==> [resin]^-Ca²⁺[resin]^-_(s) + 2H⁺_(aq)

 or 2[resin]^-Na⁺_(s) + Mg²⁺_(aq) ==> [resin]^-Mg²⁺[resin]^-_(s) + 2Na⁺_(aq) etc.

(5) Scum formation with hard water:

On mixing hard water with soaps made from the sodium salts of fatty acids, insoluble calcium or magnesium salts of the soap are formed as a grey precipitate ...

CaSO_4(aq) + 2C₁₇H₃₅COONa_(aq) ==> (C₁₇H₃₅COO)₂Ca_{(s for scum!)} + Na₂SO_4(aq)

or more simply ionically: Ca²⁺_(aq) + 2C₁₇H₃₅COO^-_(aq) ==> (C₁₇H₃₅COO^-)₂Ca²⁺_(s)

the spectator ions are SO₄^2- and Na⁺

Revision notes on writing equations in chemistry, how to balance chemical equations, how to read and write formulae, word equations, balancing symbol equations, how to write and balance ionic equations, how to work out a formula from valencies, how to work out formula from the charges on the ions,  help when revising for AQA GCSE chemistry A level chemistry, Edexcel GCSE chemistry A level chemistry, OCR GCSE gateway science chemistry, OCR 21st century science chemistry, OCR A level chemistry, Salters A level chemistry, GCSE 9-1 chemistry,  examinations.