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Answer: For the reaction shown, calculate the theoretical

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 50P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 50P

PROBLEM 50P

For the reaction shown, calculate the theoretical yield of the product in moles for each of the initial quantities of reactants.

Ti(s) + 2 Cl2(g) → TiCl4(s)

(a) 2 mol Ti; 2 mol Cl2

(b) 5 mol Ti; 9 mol Cl2

(c) 0.483 mol Ti; 0.911 mol Cl2

(d) 12.4 mol Ti; 15.8 mol Cl2

Step-by-Step Solution:
Step 1 of 3

Solution 50P

Here, we are going to calculate the theoretical yield of product in moles.

Therefore, for that we have to calculate the limiting reactant or reagent for the reaction.

The reactant which is completely consumed in a chemical reaction and limits the product formation is called limiting reagent.

The amount of product formed based on the limiting reagent is called theoretical yield.

Ti(s) + 2 Cl2(g) → TiCl4(s)

2 mol Ti; 2 mol Cl2

Here, conversion is mole to mole

Mol Ti(s) = mol TiCl4(s) = mol Cl2 (g)

Therefore,

2 mol Ti x = 2.0 mol TiCl4(s)

Similarly,  2 mol Cl2 x = 1.0 mol TiCl4(s)

Thus, we know least amount of product formation determines the limiting reagent.

Hence, the limiting reactant is Cl2 (g)

Therefore, the theoretical yield of the product (TiCl4(s)) from the reaction having (a) is 1.0 mole.

(b) 5 mol Ti; 9 mol Cl2

Ti(s) + 2 Cl2(g) → TiCl4(s)

Here, conversion is mole to mole

Mol Ti(s) = mol TiCl4(s) = mol Cl2 (g)

Therefore,

5 mol Ti x = 5.0 mol TiCl4(s)

Similarly,  9 mol Cl2 x = 4.5 mol TiCl4(s)

Thus, we know least amount of product formation...

Step 2 of 3

Chapter 8, Problem 50P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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