Writing and Balancing Chemical Equations

Writing and Balancing Chemical Equations

• A reaction equation can describe both physical and chemical changes.
• The starting materials (reactants) are on the left  while the final products are on the right.
• The following combustion reaction can be verbally stated as follows.

Chemical Equation for Physcial Change

An example of a chemical reaction showing a physical change is as folllows, which represents the melting of Ice (i.e. water in the solid state) to liquid water.

H₂O_(s)  ➞ H₂O_(l)

In this equation, the (s) indicates the water is in the gas phase and (l) is for the liquid phase.  We will also encounter (s) to indicaste solids and (aq) to represent aqueous solutions. 

Chemical equations for Chemical Change

An example of a chemcial reaction for a chemical change is the combustion of methane (CH₄).

CH₄ + O₂  ➞  CO₂ + H₂O

Verbally we would say "One molecule (or mole) of CH₄ reacts with 2 molecules (or moles) of O₂ to produce one molecule (or mole) of CO₂ and two molecules of H₂O (water)".

The ratios expressed for this particular reaction hold true no matter how many moles one startes with … if you double the moles of 1 component, all other must be doubled to hold true.

The fixed ratio of this particular reaction allows us to predict how many moles (and by extension, grams) of product we could make if we start with a specific mass (or moles) of a reactant.

Balanced Chemical Reactions

The following reaction is balanced.

CH₄ + O₂  ➞  CO₂ + H₂O

This means that the numbers and types of elements on the left are equal to the numbers and types of elements on the right. This must always be true to obey the Conservation of Mass.

Unbalanced Chemical Reactions

The following chemical reaction is unbalanced.  This reaction represents the electrolysis (i.e. using electricity to split a water molecule in to hydrogen (H2) and oxygen (O2)).

H₂O  ➞  O₂ + H₂

2+Is the following balanced? Na(s) + H2O(l)  NaOH(aq) + H2(g) No. While Na and O are equal on both sides, there are 2 H on the left and 3 H on the right. What do we do? Hint: Easiest to create even numbers of atoms. Na(s) + H2O(l)  2 NaOH(aq) + H2(g) Since NaOH has all three element types, making them even in number (2) allows us to alter the other compounds more easily. 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)

Your goal is to balance this equation so equal numbers of each atom are on each side.

How to Balance: The Trial-and-Error Method

This technique is know as the "Trial and Error" method.

1. Write the correct formulas for reactants and products.

2. Count the atoms of each element on both sides.
3. Use coefficients to balance elements one at a time. Tip: Leave Oxygen and Hydrogen for last.

Example Problem 1

Is the following balanced?
Na(s) + H₂O(l)  ➞ NaOH(aq) + H₂(g)

No. While Na and O are equal on both sides, there are 2 H on the left and 3 H on the right. What do we do? Hint: Easiest to create even numbers of atoms.

Na(s) + H₂O(l) ➞ 2 NaOH(aq) + H₂(g)

Since NaOH has all three element types, making them even in number (2) allows us to alter the other compounds more easily.

2 Na(s) + 2 H₂O(l) ➞ 2 NaOH(aq) + H₂(g)

Example Problem 2

Is the following equation balanced?

NH₄NO₃ ➞ N₂ + O₂ + H₂O

Let’s double the H2O
NH₄NO₃ ➞ N₂ + O₂ + 2 H₂O

Now there are 3 O on left, 4 on the right. Since the left is odd, let’s double the reactant.
2 NH₄NO₃ ➞ N₂ + O₂ + 2 H₂O

Now there are 6 O, 4 N, and 8 H on the left; 4 O, 2 N, and 4 H on the right. At least they’re all even. First, let’s fix nitrogen…
2 NH₄NO₃ ➞ 2 N₂ + O₂ + 2 H₂O

Now we have 6 O, 4 N, and 8 H on the left; 4 O, 4 N, and 4 H on the right.

We need two more O on the right and twice the number of H on the right. Let’s alter H2O again, since it has both elements in it:
2 NH₄NO₃ ➞ 2 N₂ + O₂ + 4 H₂O

Example Problem 3

Sometimes, it may be easier to use a temporary fraction in the balancing process. If we burn butane in oxygen, we create water and carbon dioxide (same with any hydrocarbon fuel).
C₄H₁₀ + O₂ ➞ CO₂ + H₂O

We can balance the carbons and hydrogens first easily on the right, since they’re in separate compounds.
C₄H₁₀ + O₂ ➞ 4 CO₂ + 5 H₂O

Now carbons (4) and hydrogens (10) are balanced. What about oxygen?

Since oxygen stands alone in O2, but we have an even and odd number in CO2 and H2O, let’s use a temporary fraction.

C₄H₁₀ + O₂ ➞ 4 CO₂ + 5 H₂O

Since we need 13 O atoms (those on the right) but we have O₂, by multiplying by 13/2 … we get 13 O atoms on the left.

C₄H₁₀ + 13/2 O₂ ➞ 4 CO₂ + 5 H₂O

It’s now balanced but we don’t like fractions in our equations. Since this is balanced, we can multiply everything by ‘2’ to get rid of the fraction:
2 C₄H₁₀ + 13 O₂ ➞ 8 CO₂ + 10 H₂O

Example Problem 4

When balancing reactions that contain polyatomic ions, we can often (not always) treat the polyatomic ion as one unit

CaCl₂(aq) + AgNO₃(aq) ➞ Ca(NO₃)₂(aq) + AgCl(s)

Because nitrate (NO₃^2-) is a polyatomic ion that does not change to something else in the reaction, we can balance it as a whole … we don’t need to balance just the N or O atoms as previously shown.

Full Ionic equation

So let’s balance the nitrates

CaCl₂(aq) + 2 AgNO₃(aq) ➞ Ca(NO₃)₂(aq) + AgCl(s)

Which now imbalances silver (Ag). Fix that by doubling AgCl…
CaCl₂(aq) + 2 AgNO₃(aq) ➞ Ca(NO₃)₂(aq) + 2 AgCl(s)

That resolved the unbalanced Ag and the Cl.

Example Physcial Change Reactions

We sometimes write reactions that don’t quite seem like reactions but rather show physical changes:

Adding heat to ice produces liquid: H₂O(s) + Δ ➞ H₂O(l)

Dissolving salt in water: NaCl(s) ➞ Na⁺ (aq) + Cl^- (aq)

Sometimes, a chemical reaction can occur where only some of the atoms or ions create products while others remain unchanged. For example, if we mix a solution of CaCl2(aq) with a solution of AgNO3(aq) , the reaction can be written as:

CaCl₂(aq) + 2 AgNO₃(aq) ➞ 2 AgCl(s) + Ca(NO₃)₂(aq)

Since several of these reagents are in the aqueous form, the ionic salts are really present as their individual ions in water, and we can expand this to:

Ca²⁺ (aq) + 2 Cl^- (aq) + 2 Ag+(aq) + 2 NO₃^- (aq) ➞ 2 AgCl(s) + Ca²⁺ (aq) + 2 NO₃^- (aq)     Full Ionic equation

Note that above, the Ca²⁺ (aq) and 2 NO₃^- (aq) appear unchanged on both sides of the reaction. We call these spectator ions, they start on the reactants side paired with other ions which then dissolve. But after the reaction occurs to create AgCl(s), they remain dissolved and unchanged on the products side.

So, the actual chemical reaction of interest here is just

2 Cl^- (aq) + 2 Ag⁺ (aq) ➞ 2 AgCl(s)    Net Ionic Equation

The first reaction is the full ionic reaction, while the second is called the net ionic reaction. We can further simplify be removing all of the ‘2’s.

Mixing two aqueous reactants together to form a solid product is called a precipitation reaction. The above example can also be called a double displacement reaction, where the reactant cations and anions switch with one another to form the products.

Final Notes

Any time a reaction occurs where an aqueous component is found on both reactant and product side, unchanged, it can be removed as a spectator ion. But if a new compound is formed in the solid, liquid, or gas phase, it may not be removed.

CaCO₃(aq) + 2 HCl(aq) ➞ CaCl₂(aq) + CO₂(g) + H₂O(l)

Ca²⁺(aq) + CO₃^2-(aq) + 2H⁺(aq) + 2Cl^- (aq) ➞ Ca2+(aq) + 2Cl^- (aq) + CO₂(g) + H₂O(l)

Here, Ca²⁺(aq) and 2Cl^-(aq) are the spectator ions; the net ionic is

CO₃^2-(aq) + 2H⁺(aq) ➞ CO₂(g) + H₂O(l)