Exercise 8.50 For the reaction shown, compute the theoretical yield of the produ

Question

Exercise 8.50

For the reaction shown, compute the theoretical yield of the product in moles for each of the initial quantities of reactants.
Ti(s)+2Cl2(g)TiCl4(s)

Part A

2 molTi; 2 molCl2

Express your answer using two significant figures.

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Part B

5 molTi; 9 molCl2

Express your answer using two significant figures.

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Part C

0.483 molTi; 0.911 molCl2

Express your answer using three significant figures.

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Part D

12.4 molTi; 15.8 molCl2

Express your answer using three significant figures.

Explanation / Answer

Ti+2Cl2 ---> TiCl4

Part A.

2 moles of Ti requires 4 moles of Cl2 to completely react, but there are only 2 moles of Cl2 and Cl2 is the limiting reactant

2 moles of Cl2 gives 1 mole of TiCl4

Part B.

5 moles require 10 moles of Cl2 to completely react. But there are only 9 moles. Hence, Ti is excess and Cl2 is limiting reactant

9 moles of Cl2 requires 4.5 moles of Ti and gives rise to 4.5 moles of TiCl4

Part C

0.483 moles completely react if 2*0.483= 0.966 moles of Cl2 is there .But there are only 0.911 moles of Cl2. Hence Cl2 is limiting

0.911 moles require 0.911/2=0.455 moles of Ti and produces 0.455 moles of TiCl4.

Part D

12.4 moles of Ti requires 2*12.4= 24.8 moles of Cl2. But there are only 15.8 moles of Cl2. Hence Cl2 is limiting reactant and consumed 15.8/2 =7.9 moles of Ti and 7.9 moles of TiCl4.

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