You mix 7.6 g of calcium and 81.1 g of iodine to form calcium iodide. How many moles of calcium iodide will be produced?

Molar mass of Ca = 40.08 g/mol
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol
Molar mass of CaI₂ = (40.08 + 126.9×2) g/mol = 293.9 g/mol

Ca + I₂ → CaI₂
OR: Mole ratio Ca : I₂ : CaI₂ = 1 : 1 : 1

Initial no. of moles of Ca = (7.6 g) / (40.08 g/mol) = 0.19 mol
Initial no. of moles of I₂ = (81.1 g) / (253.8 g/mol) = 0.32 mol > 0.19 mol
Hence, I₂ is in excess, and Ca is the limiting reactant (completely reacts).

No. of moles of Ca reacted = 0.19 mol
No. of moles of CaI₂ produced = 0.19 mol