Let's balance the given chemical equations and classify them.
a) FeCl₃(aq) + KOH(aq) → Fe(OH)₃(s) + KCl(aq)
Balancing:
- We have 1 Fe on both sides.
- We have 3 Cl from FeCl₃ on the left and KCl on the right. To balance Cl, we need 3 KCl.
- Thus, we also need 3 KOH to provide 3 OH⁻ for Fe(OH)₃.
Balanced Equation: \[ \text{FeCl}_3(aq) + 3 \text{KOH}(aq) \rightarrow \text{Fe(OH)}_3(s) + 3 \text{KCl}(aq) \]
Classification: This reaction is a double displacement (or metathesis) reaction, as it involves the exchange of ions between reactants.
b) CH₄ + H₂O → CO + H₂
Balancing:
- We have 1 C on both sides.
- In the reactants, we have 4 H in CH₄, and on the products side, we have 2 H in H₂. To balance H, we need 2 H₂ on the right side (which gives us 4 H).
- We also need to balance the O. In the reactants, we have 1 O from H₂O, and we have 1 O in CO.
Balanced Equation: \[ \text{CH}_4 + \text{H}_2\text{O} \rightarrow \text{CO} + 3 \text{H}_2 \]
Classification: This reaction is a combustion reaction (specifically, partial combustion, since it produces CO instead of CO₂), as it involves a hydrocarbon reacting with water to produce carbon monoxide and hydrogen gas.
Final Summary:
- a) Balanced: \[ \text{FeCl}_3(aq) + 3 \text{KOH}(aq) \rightarrow \text{Fe(OH)}_3(s) + 3 \text{KCl}(aq) \] (Double displacement)
- b) Balanced: \[ \text{CH}_4 + \text{H}_2\text{O} \rightarrow \text{CO} + 3 \text{H}_2 \] (Partial combustion)