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For the reaction shown, calculate how many grams of oxygen

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

Solution for problem 32P 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 32P

PROBLEM 32P

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

(a) 2.72 g KClO3

(b) 0.361 g KClO3

(c) 83.6 kg KClO3

(d) 22.4 mg KClO3

Step-by-Step Solution:
Step 1 of 3

Solution 32P

Here, we are going to calculate the mass of oxygen.

2 KClO3(s) → 2KCl (s) + 3O2(g)

(a) 2.72 g KClO3

g KClO3 = mol KClO3 = g O2 = mol O2

Thus,

Molar mass of  KClO3 = 122.55 g/mol

Molar mass of O2 =32 g/mol

2.72 g KClO3 x  x  x

= = 1.065 g of O2

Thus , 2.72 g KClO3  forms =  1.065 g of O2

(b) 0.361 g KClO3

g KClO3 = mol KClO3 = g O2 = mol O2

Thus,

Molar mass of  KClO3 = 122.55 g/mol

Molar mass of O2 =32 g/mol

0.361 g KClO3 x  x  x

= = 0.141 g of O2

Thus ,

0.361 g KClO3  forms =  0.141 g of O2

(c) 83.6 kg KClO3

g KClO3 = mol KClO3 = g O2 = mol O2

Thus,

Molar mass of  KClO3 = 122.55 g/mol

Molar mass of O2 =32...

Step 2 of 3

Chapter 8, Problem 32P is Solved
Step 3 of 3

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

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For the reaction shown, calculate how many grams of oxygen

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