Question

For the reaction shown, compute the theoretical yield of the product in moles for each of the initial quanities of reactants. Ti(s)+ 2Cl2(g)=TiCl4(s) A) 2 mol Ti, 2 mol Cl2 express your answer using two significant figures. B) 5 mol Ti, 9 mol Cl2 express your answer using two significant figures. C) 0.483 mol Ti, 0.911 mol Cl2 express your answer using three significant figures. D) 12.4 mol Ti, 15.8 mol Cl2 express your answer using three significant figures

Answer 1

Ti (s) + 2Cl2 (g) = TiCl4 (s)

A) we had 2 mol Ti and 2mol Cl2

as per reaction 2Cl2 reacts per 1Ti ,

hence Cl2 needed per 2 Ti moles = ( 2/1) Ti moles = ( 2/1) x 2 = 4 mol

since we had only 2Cl2 which is less than required , Cl2 is limiting reagent

Product moles depend on limiting reagent moles

Hence TiCl4 moles = ( 1/2) Cl2 moles = ( 1/2) ( 2) = 1.0 mol   is our thoeretical yield

B) Cl2 moles needed per 5 mol Ti = ( 2/1) x 5 = 10 mol

since we have only 9 mol Cl2 it sis limiting reagent

TiCl4 moles = 9 1/2) Cl2 moels = ( 1/2) ( 9) = 4.5 mol is our theoretical yield

C) Cl2 mol needed per 0.483 mol Ti = ( 2/1 ) ( 0.483) = 0.966

but since we have only 0.911 mol Cl2 , Cl2 is limiting reagent

TiCl4 moles = ( 1/2) ( 0.911) = 0.456 mol is our thoeretical yleld

D) Cl2 mol needed per 12.4 mol Ti = ( 2/1) ( 12.4) = 24.8 mol

since Cl2 is relatively less than required it is limiting reagent

TiCl4 moles = ( 1/2) Cl2 moles = ( 1/2) ( 15.8) = 7.90 mol is our theoretical yield