03 Chemical Reactions and Stoichiometry

Writing and Balancing Equations

A chemical equation represents a reaction by placing reactants on the left and products on the right. Formulas identify the substances, plus signs separate multiple substances, and an arrow shows the direction of the reaction. State symbols can indicate whether each substance is a solid, liquid, gas, or dissolved in water.

A chemical equation must obey conservation of matter: each element must have the same number of atoms on both sides. Balance an equation by adjusting its coefficients, not its subscripts. Subscripts are part of each substance’s formula, so changing one would change the substance itself. Reduce coefficients to the smallest whole-number ratio when possible.

For example, the unbalanced formation of water can be written as:

Balancing oxygen and then hydrogen gives:

There are four hydrogen atoms and two oxygen atoms on each side. In this balanced chemical equation, the coefficients also give the ratio in which the substances react and form.

Takeaway: Balance atoms by changing coefficients while preserving each chemical formula.

Recognizing Reaction Types

Reaction categories describe common patterns. Some reactions fit more than one category, so the labels are useful ways to recognize what is happening rather than mutually exclusive boxes.

• Synthesis (combination): Simpler substances form a more complex product. For example, $\mathrm{2H_2 + O_2 \longrightarrow 2H_2O}$.

• Decomposition: One compound breaks into simpler substances. For example, $\mathrm{2H_2O_2 \longrightarrow 2H_2O + O_2}$.

• Single displacement: One element replaces another element in a compound. For example, $\mathrm{Zn + 2HCl \longrightarrow ZnCl_2 + H_2}$.

• Double displacement: Ions exchange partners, often producing a precipitate, water, or a gas. For example, $\mathrm{AgNO_3(aq) + NaCl(aq) \longrightarrow AgCl(s) + NaNO_3(aq)}$.

• Combustion: A substance reacts with oxygen. Combustion of a hydrocarbon commonly produces carbon dioxide and water: $\mathrm{CH_4 + 2O_2 \longrightarrow CO_2 + 2H_2O}$.

• Acid–base neutralization: An acid and a base react, commonly producing a salt and water: $\mathrm{HCl + NaOH \longrightarrow NaCl + H_2O}$.

Other useful classifications include precipitation, acid–base, and oxidation–reduction reactions. In an oxidation–reduction (redox) reaction, electrons are transferred and oxidation states change.

Takeaway: Identify a reaction pattern by comparing the arrangement and types of reactants and products.

Using Mole Ratios

Stoichiometry uses the coefficients in a balanced equation to relate the amounts of substances reacting and forming. These coefficient ratios are mole ratios, not mass ratios. For example, in $\mathrm{2Al + 3Cl_2 \longrightarrow 2AlCl_3}$, the ratio of aluminum chloride to aluminum is $\frac{2\ \mathrm{mol\ AlCl_3}}{2\ \mathrm{mol\ Al}}$, which simplifies to $1:1$.

To convert a mass of one substance into a mass of another, follow this sequence:

1. Convert the given mass to moles using its molar mass.

2. Use the balanced equation’s mole ratio to find moles of the desired substance.

3. Convert those moles to the requested unit, such as grams.

In shorthand, the path is $\mathrm{grams\ A \longrightarrow moles\ A \longrightarrow moles\ B \longrightarrow grams\ B}$.

For the aluminum reaction, if chlorine is available in excess, $0.50\ \mathrm{mol}$ of aluminum can produce $0.50\ \mathrm{mol}$ of aluminum chloride because the mole ratio is $1:1$.

Takeaway: Balance the equation first, then use its coefficients as mole ratios between substances.

Limiting Reactants and Reaction Yields

When reactants are supplied in amounts that do not match the balanced equation’s mole ratio, one reactant is consumed first. The Limiting reactant determines the maximum amount of product; the other reactant is in excess.

A reliable way to identify the limiting reactant is to calculate how much product each reactant could form. The reactant that produces the smaller amount is limiting.

For example, consider $\mathrm{2H_2 + O_2 \longrightarrow 2H_2O}$. Starting with $5.0\ \mathrm{mol}$ of hydrogen and $2.0\ \mathrm{mol}$ of oxygen:

• The hydrogen could form $5.0\ \mathrm{mol}$ of water.

• The oxygen could form only $4.0\ \mathrm{mol}$ of water.

Therefore, oxygen is the limiting reactant, and the theoretical yield is $4.0\ \mathrm{mol}$ of water. Hydrogen is in excess, with $1.0\ \mathrm{mol}$ left unreacted.

The Theoretical yield is the maximum product predicted from the limiting reactant. The actual yield is the amount obtained experimentally. Side reactions, incomplete reaction, and losses during collection can make the actual yield smaller than the theoretical yield.

Percent yield compares these two amounts:

If the actual yield is $3.2\ \mathrm{mol}$ of water and the theoretical yield is $4.0\ \mathrm{mol}$, then:

Use the same units for actual and theoretical yield so their ratio is meaningful.

Takeaway: The limiting reactant sets the theoretical yield; comparing actual yield with theoretical yield gives percent yield.

Stoichiometry in Solutions

Molarity describes the amount of dissolved substance per volume of solution. It relates the amount in moles, $n$, to volume in liters, $V$:

Here, $M$ is measured in moles per liter. Convert solution volumes to liters before using the relationship. Once the known solution amount is expressed in moles, apply the balanced equation’s mole ratio as in any other stoichiometry problem.

Example: Find the volume of $0.100\ \mathrm{M}$ sodium hydroxide needed to neutralize $25.0\ \mathrm{mL}$ of $0.200\ \mathrm{M}$ hydrochloric acid.

1. The balanced equation is $\mathrm{HCl(aq) + NaOH(aq) \longrightarrow NaCl(aq) + H_2O(l)}$.

2. Convert the acid volume to liters and calculate its amount: $0.200\ \mathrm{mol\,L^{-1}} \times 0.0250\ \mathrm{L} = 0.00500\ \mathrm{mol\ HCl}$.

3. The equation gives a $1:1$ mole ratio, so $0.00500\ \mathrm{mol}$ of sodium hydroxide is required.

4. Calculate the required base volume: $V = \frac{n}{M} = \frac{0.00500\ \mathrm{mol}}{0.100\ \mathrm{mol\,L^{-1}}} = 0.0500\ \mathrm{L} = 50.0\ \mathrm{mL}$.

The same sequence applies to other reactions in solution: balance the equation, convert known solution quantities to moles, apply the mole ratio, and convert to the requested quantity.

Takeaway: Use concentration and volume to find moles before applying the balanced equation’s mole ratio.