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Consider the reaction: A sample of CaCO3 is placed into a sealed 0.500-L container and

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

Solution for problem 103P Chapter 15

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 103P

Consider the reaction:

\(\mathrm{CaCO}_{3} \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

A sample of \(\mathrm{CaCO}_{3}\) is placed into a sealed 0.500-L container and heated to 550 K at which the equilibrium constant is \(4.1^{*} 10^{-4}\). When the reaction has come to equilibrium, what mass of solid CaO is in the container? (Assume that the sample of \(\mathrm{CaCO}_{3}\) was large enough

that equilibrium could be achieved.)

Equation Transcription:

CaCO3

4.1 * 10-4

Text Transcription:

CaCO_3 Harpoon CaO(s) + CO_2(g)

CaCO_3

4.1 times 10^-4

Step-by-Step Solution:

Solution:

Step 1

The equilibrium constant is

Keq =

It is known that the equilibrium constants for pure solids and liquids are zero. So that [CaCO3(s)] and [CaO(s)] will be equal to zero. The above equation can be written as,

Keq = [CO2]

Here it has been given that Keq = 4.1 × 10−4

Thus Keq = [CO2] = 4.1× 10−4 = 0.00041M

The mole of CaO = 0.5 L× 0.00041 mol/L= 0.000205 mol

Molar mass of CaO = 56.07g/mol

Thus mass of CaO in the container = 56.07 g/mol  × 0.000205 mol = 0.0115g

Thus the container have 0.0115 g of CaO.

Step 2 of 1

Chapter 15, Problem 103P is Solved
Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Consider the reaction: A sample of CaCO3 is placed into a sealed 0.500-L container and