Introduction to Chemical Reactions, Stoichiometry, and Solutions

Mole Ratios

Once an equation is balanced, the coefficients provide the mole ratios. These ratios allow us to predict how much product will form from a given amount of reactant. For example, in the reaction: $4\text{NH}_3(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{NO}(\text{g}) + 6\text{H}_2\text{O}(\text{l})$ The ratio of $\text{NH}_3$ to $\text{O}_2$ is $4:5$. If you have 8.00 moles of ammonia, you can calculate the required oxygen using this ratio.

Limiting and Excess Reactants

In real-world laboratory settings, reactants are rarely present in the exact stoichiometric proportions required. This leads to two important concepts:

1. Limiting Reactant: The reactant that is completely consumed first, thereby limiting the amount of product that can be formed. It is the reactant that produces the lowest mass of product.
2. Excess Reactant: The reactant that remains in the container after the reaction has gone to completion.

The diagram above illustrates how, in a closed system, the total number of atoms remains constant even as they rearrange into new molecules, demonstrating mass conservation.

Combustion and Gas Behavior

A common type of reaction is the combustion reaction, such as the burning of propane ($\text{C}_3\text{H}_8$) in oxygen to produce carbon dioxide and water vapor. When these reactions involve gases, we often use the Ideal Gas Law to predict their behavior: $\text{PV} = \text{nRT}$ Where $\text{P}$ is pressure, $\text{V}$ is volume, $\text{n}$ is moles, $\text{R}$ is the ideal gas constant ($0.08206 \text{ L}\cdot\text{atm}/\text{mol}\cdot\text{K}$), and $\text{T}$ is temperature in Kelvin.

When working with gases, we frequently refer to Standard Temperature and Pressure (STP), defined as $1.0 \text{ atm}$ and $273 \text{ K}$. At STP, one mole of an ideal gas occupies $22.4 \text{ L}$. In most chemistry problems, you should assume ideal behavior for gases unless explicitly told otherwise.

Solubility and Precipitation Reactions

When two aqueous solutions are mixed, they may undergo a double-replacement reaction that results in the formation of a precipitate—an insoluble solid that "falls out" of the solution.

Solubility Rules

To predict whether a precipitate will form, chemists use solubility rules:

• Highly Soluble: Most nitrate salts, and most salts containing sodium ($\text{Na}^+$), potassium ($\text{K}^+$), or ammonium ($\text{NH}_4^+$) cations.
• Often Insoluble: Most hydroxide, sulfide, carbonate, and phosphate salts, unless they are paired with $\text{Na}^+$, $\text{K}^+$, or $\text{NH}_4^+$.
• Exceptions: Most chloride salts are soluble, except for those containing silver ($\text{Ag}^+$), lead ($\text{Pb}^{2+}$), or mercury(I).

Net Ionic Equations and Spectator Ions

When a precipitate forms, we can write a net ionic equation to show only the species actually participating in the chemical change.

For example, in the reaction between silver nitrate and potassium carbonate: $\text{K}_2\text{CO}_3(\text{aq}) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Ag}_2\text{CO}_3(\text{s}) + 2\text{KNO}_3(\text{aq})$ The potassium ($\text{K}^+$) and nitrate ($\text{NO}_3^-$) ions do not participate in the formation of the solid; they remain dissolved in the water. These are called spectator ions. The net ionic equation focuses only on the formation of the precipitate: $2\text{Ag}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) \rightarrow \text{Ag}_2\text{CO}_3(\text{s})$

Molarity and Concentration

In aqueous chemistry, we express the amount of solute in a solution using molar concentration, commonly known as molarity (M). Molarity is defined as the number of moles of solute per liter of solution: $\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{Liters of solution}}$

It is important to note how concentration changes during laboratory procedures:

• Dilution: Adding water to a solution increases the total volume and decreases the concentration, but the total number of moles of solute remains the same.
• Concentration Changes: Adding more solute without adding water will increase the molarity, whereas pouring a portion of a solution into a new container does not change the concentration of the liquid being poured.

The mixing of two clear solutions can result in a sudden change in appearance, such as the formation of a solid, as illustrated in the diagram above.