Final answer:
To determine the moles of Cl2 required to react with 2.87 mol of Al atoms in the formation of aluminum chloride, we use the stoichiometric ratio 2 mol Al: 3 mol Cl2 from the balanced equation 2 Al + 3 Cl2 → 2 AlCl3, resulting in 4.305 mol Cl2.
Step-by-step explanation:
The formation of aluminum chloride from aluminum and chlorine gas can be represented by the following balanced chemical equation: 2 Al + 3 Cl2 → 2 AlCl3. To determine how many moles of Cl2 molecules are required to react with 2.87 mol of Al atoms, we can use the stoichiometric relationship given by the coefficients in the equation.
According to the equation, 2 moles of Al react with 3 moles of Cl2 to form 2 moles of AlCl3. This gives us a molar ratio of 2 mol Al: 3 mol Cl2. If we have 2.87 moles of Al, the amount of Cl2 required is given by the cross-multiplication of the molar ratio: (2.87 mol Al) × (3 mol Cl2)/(2 mol Al) = 4.305 mol Cl2. Thus, the correct answer is D. 4.305 mol.