Madison prepares solid titanium(IV) oxide by reacting 6.71 x 103 g of titanium(IV) chloride with 2.45 x 103 g of oxygen. TiCl4(g) + O2(g) --> TiO2(s) + 2Cl2(g)

Question

a) Calculate the number of moles of titanium(IV) chloride and oxygen.

b) Which one is the limiting reactant?

c) Calculate the maximum moles of titanium(IV) oxide that can form.

d) Calculate the the maximum mass of titanium(IV) oxide that can form.

e) If the percent yield of TiO₂ is 75%, what mass is actually formed?

Answer 1

TiCl₄ + O₂ ==> TiO₂ + 2Cl₂ ... balanced equation

a) mols TiCl₄ = 6.71x10³ g x 1 mol/189.7 g = 35.37 mols TiCl₄

mols O₂ = 2.45x10³ g x 1 mol / 32 g = 76.56 mols O₂

b) LIMITING REACTANT = TiCl₄ since they react in a 1:1 mol ratio and there are fewer mols of TiCl₄ than O₂

c) Maximum mols TiO₂ = 35.37 mols TiCl₄ x 1 mol TiO₂ / mol TiCl₄ = 35.4 mols TiO₂ (3 sig. figs.)

d) Maximum mass TiO₂ = 35.4 mols TiO₂ x 79.9 g/mol = 2828 g = 2830 g TiO₂ (3 sig. figs.)

e) Actual mass TiO₂ formed based on 75% yield = 0.75 x 2830 g = 2123 g = 2120 g TiO₂ (3 sig. figs.)