Problem 2 - General Chemistry: Principles and Modern Applications

Question

Balance the following equations by inspection. (a) \(\mathrm{P}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{PH}_{3}+\mathrm{P}_{4}\) (b) \(\mathrm{P}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{PCl}_{3}\) (c) \(\mathrm{FeCl}_{3}+\mathrm{H}_{2} \mathrm{S} \longrightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}+\mathrm{HCl}\) (d) \(\mathrm{Mg}_{3} \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}+\mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

For equation (a): \(P_{2}H_{4} \longrightarrow 2PH_{3}+P_{4}\). For equation (b): \(P_{4}+6Cl_{2} \longrightarrow 4PCl_{3}\). For equation (c): \(2FeCl_{3}+3H_{2}S \longrightarrow Fe_{2}S_{3}+6HCl\). For equation (d): \(Mg_{3}N_{2}+6H_{2}O \longrightarrow 3Mg(OH)_{2}+2NH_{3}\).

Step 1: Balance equation (a)

To balance this equation, start with the phosphorus (P) atoms. On the left-hand side, there are 2 P atoms while on the right-hand side there's 1 in PH3 and 4 in P4, making a total of 5. To resolve this, change the coefficient in front of PH3 to 4. Now, there are 2*4 = 8 H atoms in P2H4 on the left and 3*4 = 12 H in PH3 on the right. To fix this, add a coefficent of 2 in front of PH3, giving the balance equation as: \[P_{2}H_{4} \longrightarrow 2PH_{3}+P_{4}\]

Step 2: Balance equation (b)

Start with the phosphorus atoms. Both sides have 4 P atoms, which are already balanced. Now, see the chlorine (Cl) atoms. On the left, there are 2 Cl whereas on the right there are 3*4 = 12 Cl. Balance this by placing a 6 in front of Cl2 on the left: \[P_{4}+6Cl_{2} \longrightarrow 4PCl_{3}\]

Step 3: Balance equation (c)

Start with Iron (Fe) atoms. Place a coefficient of 2 in front of FeCl3 on the left to balance with the 2 Fe on the right. Now, see the sulfur (S) atoms. We have 1 S on the left and 3 on the right. Balance this by placing a 3 in front of H2S on the left side. Finally, balance the Hydrogen (H) and Chlorine (Cl) atoms by placing a 6 in front of HCl on the right side: \[2FeCl_{3}+3H_{2}S \longrightarrow Fe_{2}S_{3}+6HCl\]

Step 4: Balance equation (d)

Begin with magnesium (Mg) atoms. To match the 3 Mg atoms on the left with those on the right, place a coefficient of 3 in front of Mg(OH)2 on the right side. There are 2 nitrogen (N) atoms on the left and 1 on the right. Balance them by placing a 2 in front of NH3 on the right side. The Hydrogen atoms are already balanced! Lastly, balance the oxygen atoms by place a 6 in front of H2O on the left: \[Mg_{3}N_{2}+6H_{2}O \longrightarrow 3Mg(OH)_{2}+2NH_{3}\]

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the relationships between the quantities of reactants and products in a chemical reaction. It’s like a recipe, where the ingredients (reactants) combine in precise amounts to create a product. Understanding stoichiometry is essential for predicting how much of each substance will be used or produced in a reaction.

Key Aspects of Stoichiometry:

• Mole Ratios: The coefficients in a balanced chemical equation indicate the ratios of moles of reactants and products. This helps in understanding how much of each substance is needed or formed.
• Conversions: Stoichiometry often involves converting between grams, moles, and molecules to find missing quantities.
• Limiting Reactants: Sometimes, one reactant limits the amount of product that can be formed. Identifying the limiting reactant is crucial in stoichiometry calculations.

By mastering stoichiometry, you can tackle problems involving chemical reactions and predict the amounts of substances needed or produced in a lab experiment.

Chemical Reactions

A chemical reaction involves the transformation of one or more substances into different substances. This happens through the breaking and forming of chemical bonds. The substances you start with are called reactants, and the substances created are called products.

Types of Chemical Reactions:

• Synthesis: Two or more substances combine to form a single product. For example, \( A + B \rightarrow AB \).
• Decomposition: A single compound breaks down into two or more simpler substances. \( AB \rightarrow A + B \).
• Single Replacement: One element replaces another in a compound. \( A + BC \rightarrow AC + B \).
• Double Replacement: The ions of two compounds exchange places. \( AB + CD \rightarrow AD + CB \).

Chemical reactions are known for their unique properties where new compounds are formed, often resulting in energy release or absorption. Balancing these reactions ensures that atoms are neither created nor destroyed, reflecting the law of conservation of mass.

Conservation of Mass

The law of conservation of mass states that in a chemical reaction, matter is neither created nor destroyed. This means that the mass of the reactants equals the mass of the products. Balancing chemical equations is guided by this principle to ensure that the same amount of each element is present on both sides of the equation.

Importance of Conservation of Mass:

• Balancing Equations: Ensures that there are equal numbers of each type of atom on both sides of the equation.
• Predicting Products: Helps in figuring out what products will be formed and in what amounts.
• Environmental Impact: Understanding this principle aids in predicting chemical behavior in nature and industry.

By adhering to the conservation of mass, chemists can accurately model chemical processes, ensuring safety and efficiency in chemical manufacturing and research.