1. Calculate the molar mass of AlCl₃:
* Al (Aluminum) has a molar mass of 26.98 g/mol
* Cl (Chlorine) has a molar mass of 35.45 g/mol
* Molar mass of AlCl₃ = 26.98 g/mol + (3 * 35.45 g/mol) = 133.33 g/mol
2. Calculate the moles of AlCl₃:
* Moles = mass / molar mass
* Moles of AlCl₃ = 400 g / 133.33 g/mol = 3.00 mol
3. Determine the mole ratio of Cl⁻ ions to AlCl₃:
* For every 1 mole of AlCl₃, there are 3 moles of Cl⁻ ions.
4. Calculate the moles of Cl⁻ ions:
* Moles of Cl⁻ ions = (moles of AlCl₃) * (3 moles Cl⁻ / 1 mole AlCl₃)
* Moles of Cl⁻ ions = 3.00 mol * 3 = 9.00 mol
Therefore, there are 9.00 moles of chlorine ions in 400 g of AlCl₃.
