18) an instructor is planning to demonstrate the reaction below to a class as un ex of a neutralization reaction. she does a test run, using naoh pellets and 10 m hcl. the reaction mixture spatters. she concludes that this is unsafe, the reaction rate is too fast. how can she slow it down?
naoh + hcl → nacl + h₂o
a) heat the hcl
b) use 0.2 m hcl
c) add a catalyst
d) grind the naoh pellets to a powder
part 2: determine if the following reactions are balanced, as written. if not, correct them. then classify.
19) 4al + 3o₂ → 2al₂o₃
20) 2p₂o₅ → 2p + 5o₂
21) ch₄ + 2o₂ → co₂ + 2h₂o
22) 2naclo₃ → 2nacl + 3o₂
23) 3koh + co₃(po₄)₂ → 2k₃po₄ + 3co(oh)₂
challenge:
24). complete and balance the following combustion reaction

Question

Accepted Answer

# Brief Explanations:
- **Question 18**:
    - **Option A**: Heating the HCl would increase the kinetic energy of the molecules, speeding up the reaction. So this is incorrect.
    - **Option B**: Using a less concentrated HCl (0.2 M instead of 10 M) would decrease the number of H⁺ ions available for reaction. According to the collision theory, a lower concentration of reactants (in this case, HCl) leads to fewer collisions between H⁺ and OH⁻ (from NaOH) per unit time, thus slowing down the reaction rate. This is correct.
    - **Option C**: Adding a catalyst speeds up a reaction. So this is incorrect.
    - **Option D**: Grinding NaOH pellets to a powder increases the surface area of NaOH. A larger surface area allows for more contact between NaOH and HCl, speeding up the reaction. So this is incorrect.
- **Question 19**:
    - **Left - hand side (LHS)**:
        - For Al: There are 4 Al atoms ($4Al$).
        - For O: There are $3\times2 = 6$ O atoms ($3O_{2}$).
    - **Right - hand side (RHS)**:
        - For Al: In $2Al_{2}O_{3}$, the number of Al atoms is $2\times2=4$.
        - For O: In $2Al_{2}O_{3}$, the number of O atoms is $2\times3 = 6$.
    - Since the number of Al and O atoms is the same on both sides, the reaction $4Al + 3O_{2}\to2Al_{2}O_{3}$ is balanced. This is a synthesis reaction (two or more reactants combine to form a single product).
- **Question 20**:
    - **Left - hand side (LHS)**:
        - For P: In $2P_{2}O_{5}$, the number of P atoms is $2\times2 = 4$.
        - For O: In $2P_{2}O_{5}$, the number of O atoms is $2\times5=10$.
    - **Right - hand side (RHS)**:
        - For P: There are 2 P atoms ($2P$).
        - For O: There are $5\times2 = 10$ O atoms ($5O_{2}$).
    - The reaction is not balanced for P. The balanced reaction is $2P_{2}O_{5}\to4P + 5O_{2}$. This is a decomposition reaction (a single reactant breaks down into two or more products).
- **Question 21**:
    - **Left - hand side (LHS)**:
        - For C: There is 1 C atom ($CH_{4}$).
        - For H: There are 4 H atoms ($CH_{4}$).
        - For O: There are $2\times2=4$ O atoms ($2O_{2}$).
    - **Right - hand side (RHS)**:
        - For C: There is 1 C atom ($CO_{2}$).
        - For H: In $2H_{2}O$, the number of H atoms is $2\times2 = 4$.
        - For O: In $CO_{2}$ there are 2 O atoms and in $2H_{2}O$ there are 2 O atoms, so in total $2 + 2=4$ O atoms.
    - The reaction $CH_{4}+2O_{2}\to CO_{2}+2H_{2}O$ is balanced. This is a combustion reaction (a substance reacts with oxygen, usually with the release of heat and light).
- **Question 22**:
    - **Left - hand side (LHS)**:
        - For Na: In $2NaClO_{3}$, the number of Na atoms is 2.
        - For Cl: In $2NaClO_{3}$, the number of Cl atoms is 2.
        - For O: In $2NaClO_{3}$, the number of O atoms is $2\times3 = 6$.
    - **Right - hand side (RHS)**:
        - For Na: There are 2 Na atoms ($2NaCl$).
        - For Cl: There are 2 Cl atoms ($2NaCl$).
        - For O: In $3O_{2}$, the number of O atoms is $3\times2=6$.
    - The reaction $2NaClO_{3}\to2NaCl + 3O_{2}$ is balanced. This is a decomposition reaction.
- **Question 23**:
    - **Left - hand side (LHS)**:
        - For K: There are 3 K atoms ($3KOH$).
        - For O (from KOH): There are 3 O atoms ($3KOH$).
        - For H (from KOH): There are 3 H atoms ($3KOH$).
        - For Co: There are 3 Co atoms ($Co_{3}(PO_{4})_{2}$).
        - For P: There are $2\times1 = 2$ P atoms ($Co_{3}(PO_{4})_{2}$).
        - For O (from $PO_{4}$): There are $2\times4=8$ O atoms ($Co_{3}(PO_{4})_{2}$).
    - **Right - hand side (RHS)**:
        - For K: In $2K_{3}PO_{4}$, the number of K atoms is $2\times3 = 6$.
        - For O (from $K_{3}PO_{4}$): In $2K_{3}PO_{4}$, the number of O atoms is $2\times4 = 8$. In $3Co(OH)_{2}$, the number of O atoms is $3\times2=6$. So total O atoms on RHS from $K_{3}PO_{4}$ and $Co(OH)_{2}$ is $8 + 6=14$. On LHS, total O atoms is $3 + 8=11$. For H: In $3Co(OH)_{2}$, the number of H atoms is $3\times2 = 6$. For Co: There are 3 Co atoms ($3Co(OH)_{2}$). For P: There are 2 P atoms ($2K_{3}PO_{4}$).
    - The reaction is not balanced. The balanced reaction is $6KOH+Co_{3}(PO_{4})_{2}\to2K_{3}PO_{4}+3Co(OH)_{2}$. This is a double - displacement reaction (the positive and negative ions of two ionic compounds exchange places).

# Answer:
- **Question 18**: B. Use 0.2 M HCl
- **Question 19**: Balanced. Synthesis reaction
- **Question 20**: Not balanced. $2P_{2}O_{5}\to4P + 5O_{2}$. Decomposition reaction
- **Question 21**: Balanced. Combustion reaction
- **Question 22**: Balanced. Decomposition reaction
- **Question 23**: Not balanced. $6KOH+Co_{3}(PO_{4})_{2}\to2K_{3}PO_{4}+3Co(OH)_{2}$. Double - displacement reaction

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