You have to find the stoichiometric ratio between AlCl₃ and BaCl₂. The common element between them is Cl. So, the ratio of Cl in BaCl₂ to AlCl₃ is 2/3. The molar mass of AlCl₃ is 133.34 g/mol. The solution is as follows:
Mass of AlCl₃ = (6 mol BaCl₂)(2 mol Cl/1 mol BaCl₂)(1 mol AlCl₃/3 mol Cl)(133.34 g/mol) = 533.36 g AlCl₃
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
A cation (+) will be smaller than its neutral atom, while anion (-) will be larger than its neutral atom.
I hope this helped you have a good day
