Question

Complete and balance the following equations: (a) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (c) \(\mathrm{MnO}_{2}(s)+\mathrm{C}(s) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{AlP}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (e) \(\mathrm{Na}_{2} \mathrm{~S}(s)+\mathrm{HCl}(a q) \longrightarrow\)

Short answer

(a) \( Mg_3N_2(s) + 6H_2O(l) \rightarrow 3Mg(OH)_2(s) + 2NH_3(g) \) (b) \( 2C_3H_7OH(l) + \frac{9}{2}O_2(g) \rightarrow 6CO_2(g) + 7H_2O(l) \) (c) \( MnO_2(s) + C(s) \rightarrow Mn(s) + CO_2(g) \) (d) \( AlP(s) + 3H_2O(l) \rightarrow Al(OH)_3(s) + PH_3(g) \) (e) \( Na_2S(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2S(g) \)

Step-by-step solution

(a) Balancing Mg₃N₂ + H₂O

To balance this equation, first identify the number of atoms for each element in the reactants and products. Start by adding the product in the right side of the equation which is Magnesium hydroxide (Mg(OH)₂) and Ammonia (NH₃), So it should look like: Mg₃N₂(s) + H₂O(l) → Mg(OH)₂(s) + NH₃(g) Now we can balance the equation: 1. Balance the Mg atoms: There are 3 Mg atoms on the left side, so we need 3 Mg atoms on the right side. Place a 3 before Mg(OH)₂. Mg₃N₂(s) + H₂O(l) → 3Mg(OH)₂(s) + NH₃(g) 2. Balance the N atoms: There are 2 N atoms on the left side, so we need 2 N atoms on the right side. Place a 2 before NH₃. Mg₃N₂(s) + H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g) 3. Balance the H and O atoms: Now there are 6 O atoms on the right side and 6 H atoms in the Mg(OH)₂, plus 6 H atoms in NH₃. That means we need 6 H₂O molecules. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g) The balanced equation for (a) is: \( \) Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g) \( \)

(b) Balancing C₃H₇OH + O₂

First, identify the product which is the combustion of the alcohol resulting in carbon dioxide (CO₂) and water (H₂O). C₃H₇OH(l) + O₂(g) → CO₂(g) + H₂O(l) Now let's balance the equation: 1. Balance the C atoms: There are 3 C atoms on the left side, so we need 3 C atoms on the right side. Place a 3 before CO₂. C₃H₇OH(l) + O₂(g) → 3CO₂(g) + H₂O(l) 2. Balance the H atoms: There are 7 H atoms on the left side, so we need 7 H atoms on the right side. Place a 7/2 before H₂O. C₃H₇OH(l) + O₂(g) → 3CO₂(g) + (7/2)H₂O(l) To avoid fractions in our balanced equation, we can multiply the whole equation by 2: 2C₃H₇OH(l) + 2O₂(g) → 6CO₂(g) + 7H₂O(l) 3. Balance the O atoms: Now there are 8 O atoms on the left side (6 from CO₂ and 2 from H₂O) and 9 O atoms on the right side. Place a 9/2 before O₂. 2C₃H₇OH(l) + (9/2)O₂(g) → 6CO₂(g) + 7H₂O(l) Since the equation is already integer balanced, our final balanced equation for (b) is: \( \) 2C₃H₇OH(l) + (9/2)O₂(g) → 6CO₂(g) + 7H₂O(l) \( \)

(c) Balancing MnO₂ + C

First, identify the products for this reaction: manganese (Mn) and carbon dioxide (CO₂). MnO₂(s) + C(s) → Mn(s) + CO₂(g) Now let's balance the equation: 1. Balance the Mn, O, and C atoms: The reactants and products have the same numbers of Mn, O, and C atoms, so this equation is already balanced. The balanced equation for (c) is: \( \) MnO₂(s) + C(s) → Mn(s) + CO₂(g) \( \)

(d) Balancing AlP + H₂O

First, identify the products for this reaction which are aluminum hydroxide (Al(OH)₃) and phosphine (PH₃). AlP(s) + H₂O(l) → Al(OH)₃(s) + PH₃(g) Now let's balance the equation: 1. Balance the Al, P, and H atoms: The reactants have the same numbers of Al and P atoms as the products, so we only need to balance H atoms. There are 3 H atoms on the left side, so we need 3 H atoms on the right side. Place a 3 before H₂O. AlP(s) + 3H₂O(l) → Al(OH)₃(s) + PH₃(g) The balanced equation for (d) is: \( \) AlP(s) + 3H₂O(l) → Al(OH)₃(s) + PH₃(g) \( \)

(e) Balancing Na₂S + HCl

First, identify the products for this reaction: sodium chloride (NaCl) and hydrogen sulfide (H₂S). Na₂S(s) + HCl(aq) → NaCl(aq) + H₂S(g) Now let's balance the equation: 1. Balance the Na atoms: There are 2 Na atoms on the left side, so we need 2 Na atoms on the right side. Place a 2 before NaCl. Na₂S(s) + HCl(aq) → 2NaCl(aq) + H₂S(g) 2. Balance the H and Cl atoms: Now there are 2 Cl atoms on the right side and 2 H atoms in H₂S, so we need 2 HCl molecules. Na₂S(s) + 2HCl(aq) → 2NaCl(aq) + H₂S(g) The balanced equation for (e) is: \( \) Na₂S(s) + 2HCl(aq) → 2NaCl(aq) + H₂S(g) \( \)

Key concepts

Stoichiometry

Stoichiometry is like the math behind chemistry. It helps us understand the proportions of elements involved in chemical reactions. Think of stoichiometry as a recipe for a cake. Just as you need the right amounts of ingredients to bake a perfect cake, you need specific amounts of reactants to make products in a chemical reaction. Stoichiometry is essential for:

• Predicting the amounts of products formed in a reaction.
• Calculating the amount of reactants needed to produce a desired product.
• Determining the efficiency of chemical reactions through percent yield calculations.

When you look at a balanced chemical equation, stoichiometry tells you the ratio of molecules or moles that react. This can be crucial in laboratories and industrial processes where precise amounts of chemicals are required.

Balancing Reactions

Balancing reactions ensures that the same number of each type of atom appears on both sides of the equation. This is based on the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance a chemical equation, you need to:

• Count the number of atoms of each element in the reactants and the products.
• Add coefficients in front of compounds until the number of atoms of each element is the same on both sides.
• Check that the charges are balanced if dealing with ionic compounds.

For instance, when balancing the equation for water formation, H₂ + O₂ → H₂O, we need to ensure there are 2 hydrogen and 2 oxygen atoms on both sides, so the balanced form becomes 2H₂ + O₂ → 2H₂O. Balancing ensures reactions are correctly described and understood.

Chemical Reactions

Chemical reactions are processes where substances are transformed into different substances. They involve the breaking and forming of bonds, leading to new products with distinct properties from the reactants. There are several types of chemical reactions, including:

• Synthesis: Two or more substances combine to form a single compound.
• Decomposition: A compound breaks down into simpler substances.
• Single Replacement: An element in a compound is replaced by another element.
• Double Replacement: Ions in two compounds exchange places.
• Combustion: A substance reacts with oxygen, often producing heat and light.

Understanding chemical reactions is vital in everyday life, from cooking to developing new materials and medicines.

Chemical Education

Chemical education is foundational to understanding the principles that govern reactions and the behavior of matter. It provides students with the knowledge to safely conduct experiments and interpret results. A solid education in chemistry covers:

• Basic concepts like atoms, molecules, and reactions.
• Advanced topics like organic chemistry, thermodynamics, and kinetics.
• Practical laboratory skills and safety protocols.

With effective chemical education, students develop analytical and problem-solving skills, which are crucial in scientific and everyday contexts. It encourages curiosity and innovation, driving progress in fields such as environmental science, medicine, and engineering.