Problem 4 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

(a) 2Cr + 3Cl₂ → 2CrCl₃; (b) SiO₂ + 2C → Si + 2CO; (c) 3Fe + 4H₂O → Fe₃O₄ + 4H₂.

Step 1: Balance Equation (a)

We need to balance the reaction of chromium and chlorine gas forming chromium(III) chloride:\[ \mathrm{Cr}(\mathrm{s}) + \mathrm{Cl}_2(\mathrm{g}) \longrightarrow \mathrm{CrCl}_3(\mathrm{s}) \]1. Start by balancing chlorine (Cl). We have 2 chlorine molecules on the reactant side and 3 chlorine atoms in the product. To balance Cl, we can use 3 \(\mathrm{Cl}_2\) molecules, giving us 6 Cl atoms, and have 2 \(\mathrm{CrCl}_3\) to match: \[ \mathrm{Cr}(\mathrm{s}) + 3\mathrm{Cl}_2(\mathrm{g}) \longrightarrow 2\mathrm{CrCl}_3(\mathrm{s}) \]2. Now, balance chromium. We need 2 chromium atoms on the reactant side: \[ 2\mathrm{Cr}(\mathrm{s}) + 3\mathrm{Cl}_2(\mathrm{g}) \longrightarrow 2\mathrm{CrCl}_3(\mathrm{s}) \]

Step 2: Balance Equation (b)

Balancing the reaction of silicon dioxide and carbon forming silicon and carbon monoxide:\[ \mathrm{SiO}_2(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Si}(\mathrm{s}) + \mathrm{CO}(\mathrm{g}) \]1. Start with Si and balance the silicon atoms: - Both sides have 1 silicon atom, so Si is already balanced.2. Next, balance oxygen. On the reactant side, we have 2 oxygen in \(\mathrm{SiO}_2\), needing 2 \(\mathrm{CO}\) on the product side: \[ \mathrm{SiO}_2(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Si}(\mathrm{s}) + 2\mathrm{CO}(\mathrm{g}) \]3. Finally, balance carbon. To match 2 carbon monoxide, we need 2 \(\mathrm{C}\) on the reactant side: \[ \mathrm{SiO}_2(\mathrm{s}) + 2\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Si}(\mathrm{s}) + 2\mathrm{CO}(\mathrm{g}) \]

Step 3: Balance Equation (c)

Balancing the corrosion of iron:\[ \mathrm{Fe}(\mathrm{s}) + \mathrm{H}_2\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3\mathrm{O}_4(\mathrm{s}) + \mathrm{H}_2(\mathrm{g}) \]1. Start by balancing iron (Fe). We need 3 irons on the reactant side to match \(\mathrm{Fe}_3\mathrm{O}_4\): \[ 3\mathrm{Fe}(\mathrm{s}) + \mathrm{H}_2\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3\mathrm{O}_4(\mathrm{s}) + \mathrm{H}_2(\mathrm{g}) \]2. Now oxygen, 4 O in \(\mathrm{Fe}_3\mathrm{O}_4\), which needs 4 water molecules: \[ 3\mathrm{Fe}(\mathrm{s}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3\mathrm{O}_4(\mathrm{s}) + \mathrm{H}_2(\mathrm{g}) \]3. Lastly, balance hydrogen. 4 water molecules produce 8 hydrogens: \[ 3\mathrm{Fe}(\mathrm{s}) + 4\mathrm{H}_2\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3\mathrm{O}_4(\mathrm{s}) + 4\mathrm{H}_2(\mathrm{g}) \]

Key Concepts

Stoichiometry

Stoichiometry is a fancy word for a simple concept in chemistry: it’s all about the measurement of elements in compounds and reactions, ensuring balance is always achieved. Think of it like baking; you wouldn’t want more flour on one side of a cake than the other, right?
In chemical equations, stoichiometry helps us to know how much of each substance is involved in a reaction. By ensuring that the number of atoms for each element is the same on both sides of the equation, we achieve a balanced chemical equation.

• Stoichiometry uses coefficients to balance chemical equations.
• It ensures the conservation of mass and atoms in a reaction.

Simply, stoichiometry keeps everything fair, maintaining the balance of substances. Without it, chemical reactions wouldn’t reflect what happens in the real world.

Chemical Reactions

Chemical reactions are processes where substances interact to form new substances. When balancing chemical reactions, we carefully adjust the number of each type of atom on both sides of the equation like an artist finding the perfect balance in colors.
Whether combining ingredients to make a cake or reacting chemicals to produce a desired product, we must follow certain rules, making sure no atoms are lost or created along the way during the reaction.

• Every chemical reaction involves reactants transforming into products.
• Balancing reactions requires ensuring atoms of each element are conserved.

Understanding chemical reactions is fundamental to chemistry, allowing us to predict how different substances behave when they come together.

Oxidation-Reduction Reaction

Oxidation-reduction reactions, often called redox reactions, are special types of chemical reactions where electrons are transferred between substances. It's like one substance is donating sand at the beach, and another one is happily picking it up.
Typically, these reactions are characterized by the change in oxidation states, where one substance loses electrons (oxidation) and another gains electrons (reduction).
Key features of redox reactions include:

• Involvement of electron transfer between reactants.
• Change in oxidation numbers or states of the elements involved.

Understanding redox reactions is crucial for analyzing many processes, from the rusting of iron to the functioning of batteries, showcasing their integral role in both everyday life and advanced scientific applications.