Problem 13 - Chemistry and Chemical Reactivity

Question

Complete and balance the equations for the following reactions. [Assume an excess of oxygen for (d).] (a) \(\mathrm{Na}(\mathrm{s})+\mathrm{Br}_{2}(\ell) \rightarrow\) (b) \(\mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\) (c) \(\mathrm{Al}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\)

Step-by-Step Solution

Answer

(a) \( 2\mathrm{Na} + \mathrm{Br}_2 \rightarrow 2\mathrm{NaBr} \); (b) \( 2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO} \); (c) \( 2\mathrm{Al} + 3\mathrm{F}_2 \rightarrow 2\mathrm{AlF}_3 \); (d) \( \mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \).

Step 1: Complete the Equation for Reaction (a)

The unbalanced equation is \( \mathrm{Na}(\mathrm{s}) + \mathrm{Br}_2(\ell) \rightarrow \). Sodium (Na) reacts with bromine (Br₂) to form sodium bromide (NaBr) as the product. The complete equation is \( \mathrm{Na}(\mathrm{s}) + \mathrm{Br}_2(\ell) \rightarrow \mathrm{NaBr}(\mathrm{s}) \).

Step 2: Balance Reaction (a)

In order to balance the equation \( \mathrm{Na}(\mathrm{s}) + \mathrm{Br}_2(\ell) \rightarrow \mathrm{NaBr}(\mathrm{s}) \), note that 2 sodium atoms are needed to pair with each molecule of Br₂, resulting in \( 2\mathrm{Na}(\mathrm{s}) + \mathrm{Br}_2(\ell) \rightarrow 2\mathrm{NaBr}(\mathrm{s}) \).

Step 3: Complete the Equation for Reaction (b)

The unbalanced equation is \( \mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \). Magnesium reacts with oxygen to form magnesium oxide (MgO). The complete reaction is \( \mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{MgO}(\mathrm{s}) \).

Step 4: Balance Reaction (b)

To balance the equation \( \mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{MgO}(\mathrm{s}) \), notice that you need 2 magnesium atoms to react with 1 molecule of oxygen: \( 2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s}) \).

Step 5: Complete the Equation for Reaction (c)

The unbalanced equation is \( \mathrm{Al}(\mathrm{s}) + \mathrm{F}_2(\mathrm{g}) \rightarrow \). Aluminum reacts with fluorine to form aluminum fluoride (AlF₃). The complete equation is \( \mathrm{Al}(\mathrm{s}) + \mathrm{F}_2(\mathrm{g}) \rightarrow \mathrm{AlF}_3(\mathrm{s}) \).

Step 6: Balance Reaction (c)

In the equation \( \mathrm{Al}(\mathrm{s}) + \mathrm{F}_2(\mathrm{g}) \rightarrow \mathrm{AlF}_3(\mathrm{s}) \), 2 aluminum atoms and 3 fluorine molecules are needed to form 2 units of AlF₃: \( 2\mathrm{Al}(\mathrm{s}) + 3\mathrm{F}_2(\mathrm{g}) \rightarrow 2\mathrm{AlF}_3(\mathrm{s}) \).

Step 7: Complete the Equation for Reaction (d)

Given the unbalanced equation is \( \mathrm{C}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \). Carbon reacts with oxygen gas to form carbon dioxide (CO₂) since there is an excess of oxygen. The complete equation is \( \mathrm{C}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \).

Step 8: Balance Reaction (d)

The equation \( \mathrm{C}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \) is already balanced, as one carbon atom reacts with one oxygen molecule to form one molecule of carbon dioxide.

Key Concepts

Stoichiometry

Stoichiometry is a core concept in chemistry, essentially the roadmap for how a chemical reaction proceeds. It helps us understand how much of each substance is needed or produced in a reaction. Think of it like a recipe – having the perfect proportions to get the expected result.
Stoichiometry relies heavily on the molar mass of elements and compounds, allowing chemists to convert between mass and moles. It is crucial for ensuring the reactants are mixed in the correct amounts for the desired reaction to occur.
To perform stoichiometric calculations, one of the first steps is balancing the chemical equation. Balancing ensures that the same number of each type of atom is present on both sides of the equation, honoring the law of conservation of mass. For example, in the reaction between sodium and bromine to form sodium bromide, we must ensure that the number of sodium atoms and bromide ions is the same before and after the reaction.

• Reactants: Substances you start with before the reaction.
• Products: New substances formed by the reaction.
• Coefficients: Numbers in front of formulas in a balanced equation that indicate ratios of reactants and products.

By using stoichiometry in the magnesium oxide reaction, for instance, one sees clearly why two moles of magnesium are needed to react with one mole of oxygen, leading to the formation of two moles of magnesium oxide.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show what reactants start the reaction, the products formed, and their respective quantities. A basic understanding of chemical equations is vital because it sets the stage for properly interpreting and performing experiments.
A chemical equation is structured with reactants on the left, an arrow pointing to the right, and products on the right. It’s like a recipe card where every ingredient and step is laid out plainly. Balancing chemical equations is critical for accurate stoichiometric calculations and lab results.
Take, for instance, the chemical reaction of aluminum with fluorine to form aluminum fluoride. The unbalanced equation is \( \mathrm{Al} + \mathrm{F}_2 \rightarrow \mathrm{AlF}_3 \). Through balancing, we find that two aluminum atoms and three fluorine molecules are necessary to form aluminum fluoride. This completion and balancing allows the equation to accurately show that the mass and atoms are conserved through the reaction.
In some reactions, like the combustion of carbon to form carbon dioxide, the equation \( \mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \) is initially balanced. Yet, verifying it ensures that nothing was overlooked. Understanding and balancing these equations is crucial for following the law of conservation of mass, which states that no atoms are lost or gained in a reaction.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or "redox reactions," are a class of chemical reactions that involve the transfer of electrons between substances. These reactions are foundational in chemistry and involve two complementary processes: oxidation and reduction.
Oxidation is when a substance loses electrons, whereas reduction is when a substance gains electrons. Always keep in mind, "LEO the lion says GER", meaning "Loss of Electrons is Oxidation" and "Gain of Electrons is Reduction".

• Oxidizing Agent: Causes oxidation by gaining electrons and is itself reduced.
• Reducing Agent: Causes reduction by losing electrons and is itself oxidized.

In the context of balancing chemical equations, identifying changes in oxidation states helps us understand how electrons are transferred. A good example is the reaction between magnesium and oxygen, where magnesium is oxidized (loses electrons) and oxygen is reduced (gains electrons). Here, magnesium acts as the reducing agent and oxygen as the oxidizing agent.
Understanding redox reactions enhances comprehension of many natural and industrial processes, such as combustion, respiration, and corrosion. Grasping the concept of electron transfer can illuminate how energy is produced and utilized in various chemical reactions.