Pre-Laboratory Exercise: (2 points) 1. Balance the following chemical equations: a. \( \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{HNO}_{3(\mathrm{aq})} \) b. \( \mathrm{K}_{2} \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{KOH}_{(\mathrm{aq})} \) c. \( \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \) 2. Predict the products and balance the following chemical equati a. \( \mathrm{Cu}(\mathrm{OH})_{2(\mathrm{~s})}+\mathrm{HClO}_{4(\mathrm{aq})} \rightarrow \)

Let's balance the given chemical equations step by step. ### 1. Balance the following chemical equations: #### a. \( \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{HNO}_{3(\mathrm{aq})} \) To balance this equation, we need to ensure that the number of each type of atom is the same on both sides. 1. **Reactants**: - \( \mathrm{N}_{2} \mathrm{O}_{5} \) has 2 nitrogen (N) and 5 oxygen (O). - \( \mathrm{H}_{2} \mathrm{O} \) has 2 hydrogen (H) and 1 oxygen (O). 2. **Products**: - \( \mathrm{HNO}_{3} \) has 1 nitrogen (N), 1 hydrogen (H), and 3 oxygen (O). To balance the equation, we can start by balancing the nitrogen atoms. Since there are 2 nitrogen atoms in \( \mathrm{N}_{2} \mathrm{O}_{5} \), we need 2 \( \mathrm{HNO}_{3} \): \[ \mathrm{N}_{2} \mathrm{O}_{5} + \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{HNO}_{3} \] Now, let's balance the oxygen and hydrogen: - On the left, we have 5 (from \( \mathrm{N}_{2} \mathrm{O}_{5} \)) + 1 (from \( \mathrm{H}_{2} \mathrm{O} \)) = 6 oxygen. - On the right, we have 2 × 3 = 6 oxygen (from 2 \( \mathrm{HNO}_{3} \)). - For hydrogen, we have 2 on the right (from 2 \( \mathrm{HNO}_{3} \)), so we need 2 \( \mathrm{H}_{2} \mathrm{O} \) on the left. The balanced equation is: \[ \mathrm{N}_{2} \mathrm{O}_{5} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{HNO}_{3} \] #### b. \( \mathrm{K}_{2} \mathrm{O}_{(\mathrm{s})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{KOH}_{(\mathrm{aq})} \) 1. **Reactants**: - \( \mathrm{K}_{2} \mathrm{O} \) has 2 potassium (K) and 1 oxygen (O). - \( \mathrm{H}_{2} \mathrm{O} \) has 2 hydrogen (H) and 1 oxygen (O). 2. **Products**: - \( \mathrm{KOH} \) has 1 potassium (K), 1 oxygen (O), and 1 hydrogen (H). To balance the equation, we need 2 \( \mathrm{KOH} \) to match the 2 potassium atoms from \( \mathrm{K}_{2} \mathrm{O} \): \[ \mathrm{K}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH} \] Now, let's check the balance: - On the left, we have 2 K, 1 O (from \( \mathrm{K}_{2} \mathrm{O} \)), and 2 H (from \( \mathrm{H}_{2} \mathrm{O} \)). - On the right, we have 2 K (from 2 \( \mathrm{KOH} \)), 2 O (from 2 \( \mathrm{KOH} \)), and 2 H (from 2 \( \mathrm{KOH} \)). The balanced equation is: \[ \mathrm{K}_{2} \mathrm{O} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH} \] #### c. \( \mathrm{Fe}_{(\mathrm{s})}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \) 1. **Reactants**: - \( \mathrm{Fe} \) has 1 iron (Fe). - \( \mathrm{O}_{2} \) has