Question

In the reaction 2Al(s) + 3Cl₂(g) → 2AlCl₃(s), if 4.0 g of aluminum reacts with 6.0 g of chlorine gas, what is the limiting reactant?

• A. \( \text{Al} \)
• B. \( \text{Cl}_2 \)
• C. \( \text{AlCl}_3 \)
• D. None

Hint:

When determining the limiting reactant, calculate the moles of each reactant and use the stoichiometric ratio from the balanced equation. The reactant that runs out first is the limiting reactant.

Answer 1

The Correct Option is B

Step 1: Determine molar masses of reactants.- \( \text{Al} \) molar mass: 26.98 g/mol.- \( \text{Cl}_2 \) molar mass: 70.90 g/mol. Step 2: Calculate moles of each reactant.- Moles of \( \text{Al} \): \[\text{Moles of Al} = \frac{4.0 \, \text{g}}{26.98 \, \text{g/mol}} = 0.148 \, \text{mol}\]- Moles of \( \text{Cl}_2 \): \[\text{Moles of Cl}_2 = \frac{6.0 \, \text{g}}{70.90 \, \text{g/mol}} = 0.085 \, \text{mol}\] Step 3: Apply reaction stoichiometry.The balanced equation indicates a 2:3 mole ratio of \( \text{Al} \) to \( \text{Cl}_2 \). The required \( \text{Cl}_2 \) moles for complete \( \text{Al} \) reaction are:\[\text{Moles of Cl}_2 = \frac{3}{2} \times \text{Moles of Al} = \frac{3}{2} \times 0.148 = 0.222 \, \text{mol}\] Step 4: Identify the limiting reactant.With 0.085 moles of \( \text{Cl}_2 \) available, and 0.222 moles required to react with all \( \text{Al} \), \( \text{Cl}_2 \) is the limiting reactant.Answer: The limiting reactant is \( \text{Cl}_2 \).