For 2 Al + 3 Cl₂ → 2 AlCl₃, with 4 mol Al and 9 mol Cl₂, what's left unreacted (excess)?
- A
0 mol of either
- Bcheck_circle
3 mol Cl₂
- C
1 mol Cl₂
- D
1 mol Al
Explanation
To use 4 mol Al, need 6 mol Cl₂. We have 9 mol Cl₂, so 9 − 6 = 3 mol Cl₂ unreacted. Al is limiting; 4 mol Al × 3/2 = 6 mol Cl₂ consumed.