If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, cl2, to form 7.7 g of titanium (iv) chloride in a combination reaction, what is the percent yield of the product?

Ti + 2 Cl2 → TiCl4  

(3.00 g Ti) / (47.867 g Ti/mol) = 0.062674 mol Ti  

(6.00 g Cl2) / (70.9064 g Cl2/mol) = 0.084619 mol Cl2  

0.084619 mole of Cl2 would react completely with 0.084619 x (1/2) = 0.0423095 mole of Ti, but there is more Ti present than that, so Ti is in excess and Cl2 is the limiting reactant.  

(0.084619 mol Cl2) x (1 mol TiCl4 / 2 mol Cl2) x (189.679 g TiCl4/mol) = 8.025 g TiCl4 in theory  

(7.7 g) / (8.025 g) = 0.96 = 96% yield TiCl4

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