Enthalpy of formation

Calculate the standard enthalpy of formation, Δ Hof , of TiCl4 (l) in kJmol¯1, given the following information:
4 TiCl3 (s) + O2 (g) ---> 3 TiCl4 (l) + TiO2 (s) Δ Ho = -552 kJ

TiCl3 (s) + ½ Cl2 (g) ---> TiCl4 (l) Δ Ho = -83 kJ

Δ Hof of TiO2 (s) = -994 kJmol¯1

The answer is -774. But how??

Here's what you do:

Firstly you have 3 equations

4 TiCl3 (s) + O2 (g) ---> 3 TiCl4 (l) + TiO2 (s) Δ Ho = -552 kJ

TiCl3 (s) + ½ Cl2 (g) ---> TiCl4 (l) Δ Ho = -83 kJ

Ti + O2 --------> TiO2 Δ Ho = -994 kJ/mol

Rearrange them like this.

Ti + O2 --------> TiO2 Δ Ho = -994 kJ/mol
TiO2 + 3TiCl4 -----> 4TiCl3 + O2 Δ Ho = +552 see that I did here?
4TiCl3 + 2Cl2 -----> 4TiCl4 Δ Ho = -332. See what I did here?

Cancelling what is common to both sides leaves you with:

Ti + 2Cl2 ----------> TiCl4 Δ Ho - 774 kJ/mol

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