Final answer:
The limiting reactant when 2.70 g of Al and 4.05 g of Cl2 are mixed is Cl2, as it provides the smaller mole ratio compared to its stoichiometric coefficient in the balanced equation for the reaction.
Step-by-step explanation:
To determine which reactant is limiting when 2.70 g of Al and 4.05 g of Cl2 are mixed, we must first calculate the number of moles of each reactant and then use the stoichiometry of the reaction to find which reactant will form the least amount of product. The balanced chemical equation for the reaction between aluminum and chlorine gas to produce aluminum chloride is 2Al + 3Cl2 → 2AlCl3. Let's determine the moles of each reactant:
- Molar mass of Al is approximately 26.98 g/mol, so moles of Al = 2.70 g / 26.98 g/mol = 0.100 moles.
- Molar mass of Cl2 is approximately 70.90 g/mol, so moles of Cl2 = 4.05 g / 70.90 g/mol = 0.057 moles.
According to the balanced equation, 2 moles of Al reacts with 3 moles of Cl2. Now, let's compare the mole ratio of the reactants to the stoichiometric coefficients:
- For Al: 0.100 moles / 2 = 0.050
- For Cl2: 0.057 moles / 3 = 0.019
The limiting reactant is the one that gives the smaller ratio, which in this case is Cl2. Therefore, the answer is (b) Cl2.