Question

in the reaction of aqueous aluminum chloride with aqueous potassium hydroxide, the products formed are aluminum hydroxide and potassium chloride. what is the correct balanced equation for this reaction? 1) $\ce{alcl_{3}(aq) + koh(aq) -> al(oh)(aq) + kcl_{3}(s)}$ 2) $\ce{al_{3}cl(aq) + 3koh(aq) -> 3al(oh)(s) + 3kcl(aq)}$ 3) $\ce{alcl_{3}(aq) + 3koh(aq) -> al(oh)_{3}(s) + 3kcl(aq)}$ 4) $\ce{alcl_{3}(aq) + k(oh)_{3}(aq) -> al(oh)_{3}(aq) + kcl_{3}(s)}$

Explanation:

Step1: Analyze Reactants and Products Formulas

Reactants: Aluminum chloride is $\ce{AlCl_{3}}$ (aq), potassium hydroxide is $\ce{KOH}$ (aq). Products: Aluminum hydroxide is $\ce{Al(OH)_{3}}$ (s, since it's a precipitate), potassium chloride is $\ce{KCl}$ (aq).

Step2: Check Option 1

Formula errors: $\ce{Al(OH)}$ (incorrect, should be $\ce{Al(OH)_{3}}$) and $\ce{KCl_{3}}$ (incorrect, K is +1, Cl is -1, so $\ce{KCl}$). Eliminate 1.

Step3: Check Option 2

Reactant formula error: $\ce{Al_{3}Cl}$ is incorrect (should be $\ce{AlCl_{3}}$) and $\ce{Al(OH)}$ (incorrect, should be $\ce{Al(OH)_{3}}$). Eliminate 2.

Step4: Check Option 3

Reactants: $\ce{AlCl_{3}(aq) + 3KOH(aq)}$. Products: $\ce{Al(OH)_{3}(s) + 3KCl(aq)}$. Balance: Al: 1=1, Cl: 3=3 (from 3KCl), K: 3=3 (from 3KOH), OH: 3=3 (from 3KOH and $\ce{Al(OH)_{3}}$). Correct.

Step5: Check Option 4

Reactant error: $\ce{K(OH)_{3}}$ is incorrect (KOH is $\ce{KOH}$, not $\ce{K(OH)_{3}}$) and product states/ formulas: $\ce{Al(OH)_{3}(aq)}$ (should be solid) and $\ce{KCl_{3}(s)}$ (incorrect). Eliminate 4.

Answer:

1. $\ce{AlCl_{3}(aq) + 3KOH(aq) -> Al(OH)_{3}(s) + 3KCl(aq)}$