Final answer:
The student's question involves balancing chemical equations. Several examples were provided, illustrating the process of adjusting coefficients to ensure the conservation of mass. However, not all reactants were given balanced equations within the question.
Step-by-step explanation:
The questions provided are related to balancing chemical equations, which is a fundamental concept in chemistry. In such equations, the number of each type of atom on both sides of the equation must be equal, as matter cannot be created or destroyed according to the Law of Conservation of Mass. Here are the solutions for the given questions:
- (a) PC₁₅ (s) + H₂O(l) → POC₁₃ (1) + HCl(aq) To balance: PCl₅ + 4H₂O → POCl₃ + 5HCl
- (b) Cu(s) + HNO₃(aq) No balanced equation provided
- (c) H₂(g) + I₂(s) → 2HI(s) To balance: H₂ + I₂ → 2HI
- (d) Fe(s) + O₂(g) → Fe₂O₃(s) To balance: 4Fe + 3O₂ → 2Fe₂O₃
- (e) Na(s) + H₂O(l) → NaOH(aq) + H₂(g) To balance: 2Na + 2H₂O → 2NaOH + H₂
- (f) (NH₄)₂ Cr₂ O₇(s) No balanced equation provided
- (g) P₄(s) + Cl₂(g) → No balanced equation provided
- (h) PtCl₄(s) → Pt(s) + Cl₂(g) To balance: PtCl₄ → Pt + 2Cl₂
When balancing an equation, it is important to adjust coefficients to ensure the conservation of mass is observed. Careful counting of atoms on both sides of the arrow is essential, and it's helpful to start with substances that contain elements that appear in only one reactant and one product.