Chapter 4, Problem 27PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# A sample of limestone and other soil materials was heated, and the limestone decomposed to give calcium oxide and carbon dioxide.CaCO3(s) → CaO(s) + CO2(g)A 1.506-g sample of limestone-containing material gave 0.558 g of CO2, in addition to CaO, after being heated at a high temperature. What was the mass percent of CaCO3 in the original sample?

Interpretation Introduction

Interpretation:

The mass percentage of CaCO3 in the given sample of compound should be determined.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Mass percentage of reacted mass in the original sample is the ratio of mass of substance reacted to mass of whole sample taken.
• Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
Explanation

Â Balanced chemical equation for the given reaction is,

CaCO3(s)â€‰â†’â€‰CaO(s)â€‰+CO2(g)

Â The amount of CO2 was calculated from its mass. Â Because 1 mol of CaCO3 was present in the sample for each mole of CO2 isolated. Â Therefore amount of CaCO3 should be known to calculate its mass and mass percentage in the sample.

Â The molar mass of CO2 is 44.01â€‰gâ€‰/Â mol. Â The amount of this solid is,

0.558â€‰gâ€‰CO2â€‰Ã—1â€‰molâ€‰CO244.01â€‰gâ€‰CO2=â€‰0.01267â€‰molâ€‰CO2

Â 1 mol of CO2 is produced from 1 mol of CaCO3, the amount of CaCO3 in the sample must also have been 0.01267â€‰mol2.318Ã—10-3â€‰mol.

Â Therefore,

Â 0.01267â€‰molâ€‰CO2â€‰Ã—1â€‰molâ€‰CaCO31â€‰molâ€‰CO2â€‰=â€‰0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started