06/14 Chemical Reactions and Stoichiometry

Writing and Balancing Chemical Equations

Chemical reactions rearrange atoms to form new substances. A symbolic representation places the reactants on the left and the products on the right:

A balanced chemical equation must contain the same number of atoms of every element on both sides. For methane combustion, the unbalanced relationship is:

$\ce{CH4(g) + O2(g) -> CO2(g) + H2O(g)}$

Balancing it gives:

Coefficients show relative amounts in moles, molecules, or formula units. State symbols identify physical states: $\ce{(s)}$ for solid, $\ce{(l)}$ for liquid, $\ce{(g)}$ for gas, and $\ce{(aq)}$ for aqueous solution. Only coefficients may be changed during balancing; changing a subscript changes the substance itself.

A reliable balancing process is:

1. Write correct formulas for all reactants and products.

2. Count each element on both sides.

3. Adjust coefficients, beginning with elements that occur in only one substance on each side.

4. Balance hydrogen and oxygen last when they appear in several compounds.

5. Reduce the coefficients to the smallest whole-number ratio.

6. Recount every element.

Takeaway: Balance the equation before using it for any numerical calculation.

Classifying Chemical Reactions

Reaction classifications describe recurring patterns, and one reaction can fit more than one category.

• Synthesis or combination: Two or more substances form one product, following the pattern $\ce{A + B -> AB}$. An example is $\ce{2Mg(s) + O2(g) -> 2MgO(s)}$.

• Decomposition: One compound breaks into simpler substances, following $\ce{AB -> A + B}$. An example is $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$.

• Single-displacement reaction: An element replaces another element in a compound, as in $\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}$.

• Double-displacement reaction: Ions exchange partners, often producing a precipitate, gas, or water. For example, $\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}$.

• Combustion: A substance reacts rapidly with oxygen. Complete combustion of a hydrocarbon produces carbon dioxide and water, as in $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}$.

• Acid–base neutralization: An acid and a base form water and an ionic compound. The net ionic relationship is $\ce{H+(aq) + OH-(aq) -> H2O(l)}$.

• Redox reaction: Oxidation and reduction occur together through changes in oxidation number.

For ionic reactions, a molecular equation shows complete compounds, a complete ionic equation separates strong aqueous electrolytes into ions, and a net ionic equation removes spectator ions. In the precipitation of silver chloride, the net ionic equation is $\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$.

Takeaway: Identify the reaction pattern, then look for the chemical evidence that drives the reaction, such as a precipitate, gas, or water.

Stoichiometric Calculations and Mole Ratios

Stoichiometry uses the coefficients of a balanced equation to relate amounts of reactants and products. The coefficients provide exact mole ratios. For $\ce{2H2(g) + O2(g) -> 2H2O(l)}$, the relationship is $2$ mol $\ce{H2}$ to $1$ mol $\ce{O2}$ to $2$ mol $\ce{H2O}$.

The usual calculation pathway is:

Convert mass to moles using molar mass, use the coefficient ratio, and then convert to the requested unit. For example, with oxygen in excess, the mass of water formed from $5.00\ \text{g}$ of hydrogen is:

Units cancel through the calculation. The mole ratio must come from equation coefficients, not from the subscripts in the formulas.

Takeaway: Convert to moles, apply the balanced-equation ratio, and convert to the desired unit.

Limiting Reactants and Theoretical Yield

When reactants are supplied in arbitrary amounts, the limiting reactant is the one consumed first. It determines the maximum product, while the other reactant is present in excess.

For $\ce{2H2 + O2 -> 2H2O}$, consider $5.00\ \text{g}$ of $\ce{H2}$ and $10.0\ \text{g}$ of $\ce{O2}$:

1. Convert each reactant to moles:

   • $\ce{H2}: 5.00\div2.016=2.48\ \text{mol}$

   • $\ce{O2}: 10.0\div32.00=0.313\ \text{mol}$

2. Calculate the water possible from each reactant:

   • From hydrogen: $2.48\ \text{mol }\ce{H2}$ produces $2.48\ \text{mol }\ce{H2O}$.

   • From oxygen: $0.313\ \text{mol }\ce{O2}$ produces $0.625\ \text{mol }\ce{H2O}$.

3. Select the smaller possible product amount. Oxygen is limiting, so the theoretical yield is:

Do not identify the limiting reactant simply by choosing the smaller mass or smaller number of moles. Instead, calculate the product possible from each reactant. The excess reactant remaining is its initial amount minus the amount consumed; in this example, approximately $2.54\ \text{g}$ of hydrogen remains.

Takeaway: The reactant that produces the smaller amount of product is the limiting reactant.

Reaction Yields

The theoretical yield is the maximum product predicted by stoichiometry. The actual yield is the amount obtained experimentally. The actual amount may be lower because of incomplete reaction, side reactions, product loss during transfer or purification, or measurement limitations.

Percent yield is calculated by comparing these two amounts:

For an actual yield of $0.392\ \text{g}$ and a theoretical yield of $0.507\ \text{g}$:

Actual and theoretical yields must be expressed in the same units. A percent yield greater than $100\%$ commonly signals contamination, retained solvent or water, an incorrect measurement, or an incorrect theoretical calculation.

Takeaway: Use the limiting reactant to calculate the theoretical yield, then compare it with the measured actual yield.

Quantitative Chemical Analysis

Quantitative analysis determines the amount or concentration of a substance in a sample by connecting a measurement to a balanced chemical equation.

In a titration, a solution of known concentration reacts with a measured amount of an unknown solution. For hydrochloric acid and sodium hydroxide:

If $25.00\ \text{mL}$ of hydrochloric acid requires $23.40\ \text{mL}$ of $0.1000\ \text{M}$ sodium hydroxide:

The coefficient ratio is $1:1$, so the hydrochloric acid amount is also $0.002340\ \text{mol}$. Therefore:

In gravimetric analysis, an analyte is converted into a measurable solid, which is filtered, dried, and weighed. For chloride, the precipitate can be silver chloride:

The measured mass of $\ce{AgCl}$ is converted to moles and related to the original chloride amount using the equation’s mole ratio. Reliable results also require calibrated equipment, careful unit and significant-figure recording, quantitative transfer, replicate trials, and evaluation of systematic error, random error, and product loss.

Takeaway: Analytical methods turn measurements of mass, volume, or concentration into unknown quantities through balanced-equation relationships.