   Chapter 4, Problem 27PS

Chapter
Section
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

Balanced chemical equation for the given reaction is,

CaCO3(s)CaO(s)+CO2(g)

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

The molar mass of CO2 is 44.01g/ mol.  The amount of this solid is,

0.558gCO2×1molCO244.01gCO2=0.01267molCO2

1 mol of CO2 is produced from 1 mol of CaCO3, the amount of CaCO3 in the sample must also have been 0.01267mol2.318×10-3mol.

Therefore,

0.01267molCO2×1molCaCO31molCO2=0

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