What mass of titanium (IV) oxide will be produced if 85.6g of titanium (IV) chloride is reacted with 0.500 mol of water vapor?

Molar mass of TiCl₄ = (47.9 + 35.5×4) g/mol = 189.9 g/mol
Molar mass of TiO₂ = (47.9 + 16.0×2) g/mol = 79.9 g/mol

Initial moles of TiCl₄ = (85.6 g) / (189.9 g/mol) = 0.451 mol
Initial moles of H₂O = 0.500 mol

TiCl₄ + 2H₂O → TiO₂ + 4HCl
Mole ratio TiCl₄ : H₂O = 1 : 2

If H₂O completely reacts, TiCl₄ needed = (0.500 mol) × (1/2) = 0.250 mol < 0.451 mol
TiCl₄ is in excess, and thus H₂O is the limiting reactant (limiting reactant).

According to the equation, mole ratio H₂O : TiO₂ = 2 : 1
Moles of TiO₂ produced = (0.500 mol) × (1/2) = 0.250 mol
Mass of TiO₂ produced = (0.250 mol) × (79.9 g/mol) = 20.0 g