Question

In the reaction \( \text{2Al(s) + \text{3Cl}_2(g) \rightleftharpoons 2\text{AlCl}_3(s) \), 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: Calculate the molar masses of the reactants. - The molar mass of \( \text{Al} \) is 26.98 g/mol. - The molar mass of \( \text{Cl}_2 \) is 70.90 g/mol. Step 2: Calculate the 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: Use the stoichiometry of the reaction. The balanced equation shows that 2 moles of \( \text{Al} \) react with 3 moles of \( \text{Cl}_2 \). Therefore, the ratio of \( \text{Al} \) to \( \text{Cl}_2 \) is \( \frac{2}{3} \). For the moles of \( \text{Al} \) to react with \( \text{Cl}_2 \), we need: \[ \text{Moles of Cl}_2 = \frac{3}{2} \times \text{Moles of Al} = \frac{3}{2} \times 0.148 = 0.222 \, \text{mol} \] Step 4: Determine the limiting reactant. We have 0.085 moles of \( \text{Cl}_2 \), but we need 0.222 moles to fully react with the available aluminum. Therefore, chlorine gas \( \text{Cl}_2 \) is the limiting reactant. Answer: Therefore, the limiting reactant is \( \text{Cl}_2 \).