Question Detail
Question
Titanium is prepared by the reaction of titanium(IV) chloride with molten magnesium: TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(l). If 35.4 g of TiCl₄ are reacted with 11.3 g of Mg. [Ti=48, Cl=35.5, Mg=24] What is the limiting reactant for the production of Ti?
Correct Answer
Option D is the correct answer.
Detailed Explanation
Molar mass TiCl₄ = 48 + (4×35.5) = 190 g/mol. Moles TiCl₄ = 35.4/190 = 0.1863 mol. Moles Mg = 11.3/24 = 0.4708 mol. Reaction requires 2 mol Mg per 1 mol TiCl₄. Mg needed = 0.1863×2 = 0.3726 mol. Mg available = 0.4708 mol, so TiCl₄ is limiting? Wait, check: 0.4708 mol Mg would require 0.2354 mol TiCl₄, but only 0.1863 mol available. So TiCl₄ is limiting.
Hint
Compare the actual mole ratio to the stoichiometric ratio.