Chemistry 1: 2Al + 3Cl₂ → 2AlCl₃ Limiting Reactant & Theoretical Yield Problem

Exact Chemistry Midterm Problem

Solid aluminum reacts with chlorine gas according to:

> $$ 2\text{Al}(s) + 3\text{Cl}_2(g) \longrightarrow 2\text{AlCl}_3(s) $$

If $27.0\text{ g}$ of $\text{Al}$ is reacted with $100.0\text{ g}$ of $\text{Cl}_2$:

1. Identify the limiting reactant.

2. Calculate the theoretical yield of $\text{AlCl}_3$ in grams.

3. Calculate the mass of excess reactant remaining.

Step-by-Step Solution

1. Convert to Moles:

• $n_{\text{Al}} = \frac{27.0\text{ g}}{26.98\text{ g/mol}} = 1.0007\text{ mol}$
• $n_{\text{Cl}_2} = \frac{100.0\text{ g}}{70.90\text{ g/mol}} = 1.4104\text{ mol}$

1. Determine Limiting Reactant:

• Ratio needed: $\frac{3\text{ mol Cl}_2}{2\text{ mol Al}} = 1.50$
• Actual ratio: $\frac{1.4104}{1.0007} = 1.4094 < 1.50 \implies \mathbf{Cl_2\text{ is limiting}}$

1. Theoretical Yield of $\text{AlCl}_3$:

$$ n_{\text{AlCl}_3} = 1.4104\text{ mol Cl}_2 \times \frac{2}{3} = 0.9403\text{ mol} $$

$$ \text{Mass} = 0.9403\text{ mol} \times 133.34\text{ g/mol} = \mathbf{125.4\text{ g}} $$

1. Excess $\text{Al}$ Remaining:

• Consumed: $0.9403\text{ mol Al}$
• Remaining: $1.0007 - 0.9403 = 0.0604\text{ mol}$
• Mass remaining: $0.0604 \times 26.98 = \mathbf{1.63\text{ g Al}}$