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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 48P 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 48P

PROBLEM 48P

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

4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)

(a) 1 mol Cr; 1 mol O2

(b) 4 mol Cr; 2.5 mol O2

(c) 12 mol Cr; 10 mol O2

(d) 14.8 mol Cr; 10.3 mol O2

Step-by-Step Solution:
Step 1 of 3

Solution 48P

The theoretical yield is the proportion which gives the least amount of moles of product, and is called as limiting reactant,

You have 4 Cr to 3 O2. That means you need more Cr than O2.

a] 1 mol Cr has to be bigger than 1 mol O2. Since it is not, Cr is the limiting reagent.

b] 4 mol of Cr 2.5 mol of O2 

If you have 4 mol of Cr you need 3 mol of O2, the limiting reagent is O2

c] 12 mol of Cr is 3 times as large as 4. Therefore you have to multiply the 3 in front of O2 by 3. when you do you get 9 moles of O2. Since there...

Step 2 of 3

Chapter 8, Problem 48P is Solved
Step 3 of 3

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

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