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For the reaction shown, find the limiting reactant for

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

Solution for problem 47P 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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1
Problem 47P

PROBLEM 47P

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2 K(s) + Cl2(g) → 2 KCl(s)

(a) 1 mol K; 1 mol Cl2

(b) 1.8 mol K; 1 mol Cl2

(c) 2.2 mol K; 1 mol Cl2

(d) 14.6 mol K; 7.8 mol Cl2

Step-by-Step Solution:
Step 1 of 3

Solution 47P

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

2 K(s) + Cl2(g) → 2 KCl(s)

(a) 1 mol K; 1 mol Cl2

(b) 1.8 mol K; 1 mol Cl2

(c) 2.2 mol K; 1 mol Cl2

(d) 14.6 mol K; 7.8 mol Cl2

The equation tells you that 2 mol K react with 1 mol Cl2 

Ratio = 2:1

The theoretical...

Step 2 of 3

Chapter 8, Problem 47P is Solved
Step 3 of 3

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

Since the solution to 47P from 8 chapter was answered, more than 333 students have viewed the full step-by-step answer. The answer to “For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.2 K(s) + Cl2(g) ? 2 KCl(s)(a) 1 mol K; 1 mol Cl2(b) 1.8 mol K; 1 mol Cl2(c) 2.2 mol K; 1 mol Cl2(d) 14.6 mol K; 7.8 mol Cl2” is broken down into a number of easy to follow steps, and 46 words. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 47P from chapter: 8 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: mol, quantities, KCL, limiting, Find. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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