Question

aluminum metal reacts with gaseous chlorine to give solid aluminum chloride. the correct balanced equation for this reaction with all the formulas written correctly would be: 1) al(s) + cl₂(g) → alcl₂(s) 2) al(s) + cl (g)→ alcl(s) 3) al(s) + 3cl₂ (g)→ 2alcl₃(s) 4) 2al(s) + 3cl₂ (g)→ 2alcl₃(s)

Explanation:

Step1: Analyze Aluminum Chloride Formula

Aluminum (Al) has a +3 oxidation state, chlorine (Cl) has -1. So formula is $AlCl_3$, not $AlCl_2$ or $AlCl$. Eliminate options 1 and 2.

Step2: Balance the Equation

Reactants: Al (s), $Cl_2$ (g). Products: $AlCl_3$ (s).

• Al: Let's balance Al first. Suppose we have 2 Al on left (2Al), then 2 $AlCl_3$ on right (2 Al).
• Cl: 2 $AlCl_3$ has 6 Cl. $Cl_2$ has 2 Cl per molecule, so need 3 $Cl_2$ (3×2=6 Cl).

So balanced equation: $2Al(s) + 3Cl_2(g)
ightarrow 2AlCl_3(s)$. Check option 4. Option 3 has 1 Al on left, 2 on right (unbalanced Al).

Answer:

1. $2\text{Al}(s) + 3\text{Cl}_2 (g)

ightarrow 2\text{AlCl}_3(s)$