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
Answer:
adjusting the air mix by rotating the barrel and adjusting the gas with the needle valve to obtain a flame of suitable height and intensity
D is the answer!!!!!!!!!!!!!!!!
Answer:
I know someone that has the answer
Explanation:
I know someone that has the answer
