Final answer:
In the reaction between 4.79×10⁻¹ moles of Al and 1.11×10⁰ moles of Cl₂ to form AlCl₃, approximately 73.6 grams of excess Cl₂ would be left over after the reaction.
Step-by-step explanation:
To determine how many grams of the excess reactant are left over when 4.79×10⁻¹ moles of Al(s) react with 1.11×10⁰ moles of Cl₂(g), it is essential to identify the limiting reactant in the given chemical reaction.
The balanced equation for the reaction between aluminum (Al) and chlorine gas (Cl₂) to form aluminum chloride (AlCl₃) is:
2 Al + 3 Cl₂ → 2 AlCl₃
Using stoichiometry, we can calculate the proportion of moles of Al to Cl₂ required according to the equation, which is 2:3. This means we need 1.5 times as many moles of Cl₂ as moles of Al to react completely. Since Al is the excessive reactant in the given problem, we can then find the amount of Al that did not react.
First, we find the moles of Cl₂ necessary to fully react with 4.79×10⁻¹ moles of Al:
(4.79×10⁻¹ moles Al) × (3 moles Cl₂ / 2 moles Al) = approximately 7.185×10⁻¹ moles Cl₂
Since we started with 1.11×10 moles of Cl₂ which is more than 7.185×10⁻¹ moles, all 4.79×10⁻¹ moles of Al will react, and we will have leftover Cl₂. To find the remaining moles of Cl₂:
1.11×10⁰ moles Cl₂ - 7.185×10⁻¹ moles Cl₂ = approximately 1.038 moles Cl₂
The mass of the excess Cl₂ is then calculated using its molar mass (70.90 g/mol):
1.038 moles Cl₂ × 70.90 g/mol = approximately 73.6 grams of Cl₂.
Hence, after the reaction, approximately 73.6 grams of Cl₂ will be left over.