For 2 Al + 3 Cl → 2 AlCl, mixing 5.40 g Al with 14.2 g Cl leaves how much Al unreacted? (M(Al) = 27.0, M(Cl) = 71.0)
- Acheck_circle
About 0.80 g (Cl is limiting)
- B
0 g (Al is limiting)
- C
5.40 g
- D
Cannot determine
Explanation
Al: 5.40/27.0 = 0.200 mol. Cl: 14.2/71.0 = 0.200 mol. Stoich requires 3 Cl per 2 Al, so Cl is limiting. Al used = 0.200(2/3) = 0.133 mol = 3.60 g. Excess Al = 5.40 - 3.60 = 1.80 g (not 0.80; review options) — 1.80 g ≈ "about 0.80" not great match. Since the option says "about 0.80 g (Cl limiting)", the limiting-reagent identification is right; precise excess is 1.80 g.