Final answer:
The correct mole ratio for aluminum chloride to chlorine in the given chemical reaction is 3:1. By converting 123.2 g of Cl2 to moles and using the balanced equation ratio, we find that 1.159 moles of AlCl3 will be produced.
Step-by-step explanation:
The correct mole ratio for aluminum chloride (AlCl3) to chlorine (Cl2) in the chemical reaction AlCl3 + Br2 → AlBr3 + Cl2 is option C) 3:1. To determine the moles of AlCl3 produced from 123.2 g of Cl2 in the reaction, we first need to convert the mass of Cl2 to moles. The molar mass of Cl2 is 70.90 g/mol. Using this conversion factor, we can establish the mole quantity of Cl2.
Then, we use the balanced chemical equation which shows the ratio of AlCl3 to Cl2 is 2:3. This means for every 3 moles of Cl2, 2 moles of AlCl3 are produced. By calculating the mol quantity of Cl2 we have and applying this ratio, we can determine the number of moles of AlCl3 formed.
For example, if we start with 123.2 g of Cl2, we convert this to moles (123.2 g divided by 70.90 g/mol), and then use the stoichiometric ratio from the balanced equation (2 moles AlCl3 for every 3 moles Cl2) to calculate the moles of AlCl3 produced. Following this process demonstrates that 1.159 moles of AlCl3 will be produced when reacting 123.2 g of Cl2.