In the reaction 2Al + 3Cl_2 → 2AlCl_3, if 4.0 moles of Al are mixed with 6.0 moles of Cl_2, how many moles of the excess reactant remain after the completion of the reaction?

• A. 0.0 moles of Al
• B. 1.0 mole of Al
• C. 2.0 moles of Cl_2
• D. 1.0 mole of Cl_2

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2Al + 3Cl_2 \rightarrow 2AlCl_3$.
2. Identify the mole ratio between aluminum (Al) and chlorine gas (Cl_2) from the balanced equation: 2 moles of Al react with 3 moles of Cl_2.
3. Determine the limiting reactant by comparing the mole ratio of the given amounts: calculate the amount of Cl_2 needed to react with 4.0 moles of Al using the ratio \(\frac{3}{2}\), and compare it to the available 6.0 moles of Cl_2.
4. Once the limiting reactant is identified, calculate how many moles of the excess reactant are consumed by using the mole ratio from the balanced equation.
5. Subtract the moles of the excess reactant consumed from the initial moles to find the remaining moles of the excess reactant after the reaction is complete.