Problem 10 - Chemistry and Chemical Reactivity

Question

Balance the following equations: (a) \(\mathrm{Cr}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CrCl}_{3}(\mathrm{s})\) (b) \(\mathrm{SiO}_{2}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Si}(\mathrm{s})+\mathrm{CO}(\mathrm{g})\) (c) \(\mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) \(2\mathrm{Cr}+3\mathrm{Cl}_2 \rightarrow 2\mathrm{CrCl}_3\); (b) \(\mathrm{SiO}_2+\mathrm{C} \rightarrow \mathrm{Si}+2\mathrm{CO}\); (c) \(3\mathrm{Fe}+4\mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Fe}_3\mathrm{O}_4+4\mathrm{H}_2\)."

Step 1: Understand the Equation (a)

For equation (a), we have \(\mathrm{Cr}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CrCl}_{3}(\mathrm{s})\). The goal is to have the same number of each type of atom on both sides of the equation.

Step 2: Balance Chlorine in Equation (a)

There are 3 chlorine atoms in \(\mathrm{CrCl}_{3}\). To balance chlorine, we put \(\frac{3}{2}\) in front of \(\mathrm{Cl}_{2}\), as \(\mathrm{Cl}_{2}\) has two chlorine atoms. This gives us \(\mathrm{Cr}(\mathrm{s}) + \frac{3}{2}\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{CrCl}_{3}(\mathrm{s})\).

Step 3: Clear Fraction in Equation (a)

To avoid fractions, multiply the entire equation by 2: \(2\mathrm{Cr}(\mathrm{s}) + 3\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CrCl}_{3}(\mathrm{s})\). Now the equation is balanced.

Step 4: Understand the Equation (b)

For equation (b), it is \(\mathrm{SiO}_{2}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Si}(\mathrm{s}) + \mathrm{CO}(\mathrm{g})\). Balance each type of atom by comparing both sides.

Step 5: Balance Carbon and Oxygen in Equation (b)

Each \(\mathrm{SiO}_{2}\) has 2 oxygen atoms. Thus, placing a 2 in front of \(\mathrm{CO}\) gives \(\mathrm{SiO}_{2}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Si}(\mathrm{s}) + 2\mathrm{CO}(\mathrm{g})\). This also balances carbon.

Step 6: Understand the Equation (c)

Equation (c) is \(\mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g})\). Again, we match atom counts on both sides.

Step 7: Balance Iron in Equation (c)

\(\mathrm{Fe}_{3}\mathrm{O}_{4}\) contains 3 iron atoms, so we need 3 on the reactant side: \(3\mathrm{Fe}(\mathrm{s}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g})\).

Step 8: Balance Oxygen and Hydrogen in Equation (c)

There are 4 oxygen atoms in \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), so we need 4 \(\mathrm{H}_{2}\mathrm{O}\): \(3\mathrm{Fe}(\mathrm{s}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) + \mathrm{H}_{2}(\mathrm{g})\). This provides 8 hydrogen atoms for balancing \(\mathrm{H}_{2}\) on the product side.

Step 9: Final Step: Balance Hydrogen in Equation (c)

With 8 hydrogen atoms needed and \(\mathrm{H}_{2}\) contributing 2, we require 4 units: \(3\mathrm{Fe}(\mathrm{s}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s}) + 4\mathrm{H}_{2}(\mathrm{g})\). The equation is now balanced.

Key Concepts

Atoms

Atoms are the fundamental building blocks of matter. They are the smallest units of an element that retain the identity of that element. Understanding atoms is crucial in balancing chemical equations because the goal is to have the same number of each type of atom on both sides of a chemical equation.
In every balanced chemical equation:

• The quantity of each type of atom in the reactants must equal the quantity in the products.
• Atoms do not disappear or appear magically; instead, they rearrange to form different substances during the reaction.
• This conservation of atoms is a reflection of the Law of Conservation of Mass.

Balancing equations involves tracking these atoms through the reaction process.

Reactants

Reactants are the substances you start with before a chemical reaction takes place. In the balanced equations from our exercise, the reactants can be seen on the left-hand side of the equation.
For example:

• In equation (a), \(\mathrm{Cr}\) and \(\mathrm{Cl}_{2}\) are the reactants.
• In equation (b), \(\mathrm{SiO}_{2}\) and \(\mathrm{C}\) are the reactants.
• In equation (c), \(\mathrm{Fe}\) and \(\mathrm{H}_{2}\mathrm{O}\) are the reactants.

The reactants undergo chemical transformations through breaking and forming bonds, resulting in the creation of new substances known as products.

Products

Products are the substances formed as a result of a chemical reaction. They are located on the right-hand side of a chemical equation.
In our example:

• For equation (a), the product is \(\mathrm{CrCl}_{3}\).
• In equation (b), the products are \(\mathrm{Si}\) and \(2\mathrm{CO}\).
• For equation (c), the products are \(\mathrm{Fe}_{3}\mathrm{O}_{4}\) and \(4\mathrm{H}_{2}\).

The formation of products signifies that a chemical reaction has occurred, where distinct chemical species emerge from the rearrangement of atoms that were present in the reactants.

Chemical Reactions

Chemical reactions are processes where atoms and molecules interact to form new substances. In these reactions:

• Bonds between atoms in reactants break and new bonds form to create products.
• The type and number of atoms remain unchanged, adhering to the conservation of mass principle.
• Balancing a chemical reaction ensures that the equation respects the Stoichiometry, meaning the proportions of reactants and products reflect the actual number of atoms involved.

This understanding of chemical reactions is key to balancing equations, as it highlights the transformation from reactants to products without loss or gain of atoms, thus upholding the laws of chemistry.