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Multiple Choice
When 2.50 mol of Al and 3.00 mol of Cl2 combine in the reaction 2Al(s) + 3Cl2(g) → 2AlCl3(s), how many moles of the excess reactant remain at the end of the reaction?
A
1.00 mol of Cl2
B
0.50 mol of Al
C
0.50 mol of Cl2
D
1.00 mol of Al
Verified step by step guidance
1
Identify the balanced chemical equation for the reaction: 2Al(s) + 3Cl2(g) → 2AlCl3(s). This equation shows the stoichiometric relationship between the reactants and products.
Determine the limiting reactant by comparing the mole ratio of the reactants to the coefficients in the balanced equation. Calculate the moles of Al and Cl2 needed to react completely.
Calculate the moles of Al needed to react with 3.00 mol of Cl2 using the stoichiometry from the balanced equation: \( \frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2} \).
Calculate the moles of Cl2 needed to react with 2.50 mol of Al using the stoichiometry from the balanced equation: \( \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} \).
Determine the excess reactant by comparing the calculated moles needed with the initial moles available. Subtract the moles of the limiting reactant used from the initial moles of the excess reactant to find how many moles remain.