Final answer:
To balance the equation AlCl3(s) + F₂(g) → AlF₃(s) + Cl2(g), the balanced formula is 2 AlCl3(s) + 3 F₂(g) → 2 AlF₃(s) + 3 Cl2(g). The number of moles of AlCl3 formed from 123.2 g of Cl2 is calculated using molar mass and stoichiometry, resulting in about 1.158 mol of AlCl3.
Step-by-step explanation:
To balance the chemical equation AlCl3(s) + F₂(g) → AlF₃(s) + Cl2(g), we look at the number of atoms of each element on both sides of the equation. We need 2 moles of AlCl3 and 3 moles of Cl2 to be produced per reaction cycle. Therefore, the balanced equation is 2 AlCl3(s) + 3 F₂(g) → 2 AlF₃(s) + 3 Cl2(g). To determine the number of moles of AlCl3 from 123.2 g of Cl2, we use its molar mass of 70.90 g/mol.
We calculate the moles of Cl2 by dividing mass by the molar mass, which gives us 123.2 g / 70.90 g/mol ≈ 1.737 mol Cl2. From the balanced chemical equation, it is clear that 3 moles of Cl2 reacts to form 2 moles of AlCl3. So, we multiply the moles of Cl2 by the ratio of moles of AlCl3 to Cl2 from the balanced equation (2 mol AlCl3 / 3 mol Cl2), resulting in approximately 1.158 mol of AlCl3.