Chapter 4, Problem 99GQ

### 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

# What mass of lime, CaO, can be obtained by heating 125 kg of limestone that is 95.0% by mass CaCO3?CaCO3 (s) → CaO(s) + CO2(g)

Interpretation Introduction

Interpretation:

The mass of lime CaO obtained in the given reaction has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is,

Numberofmole=Mass in gramMolar mass

Mass in gram ofthesubstance=Numberofmole×Molarmass

• 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).
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• 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,

â€‚Â CaCO3(s)â€‰â†’CaO(s)â€‰+â€‰CO2(g)

From the balanced chemical equation it is clear that lime stone (CaCO3) and lime (CaO) react in 1:1â€‰ratio.

From the mass percent of lime stone its actual mass involved in this reaction can be calculated.

Thus,

â€‚Â 95% of 125 kg CaCO3 =95100 Ã—125 =118.75 kg CaCO3

The amount (moles) of CaCO3 available can be determined as follows,

â€‚Â Amount of CaCO3 =118750â€‰â€‰gâ€‰CaCO3100â€‰â€‰â€‰gâ€‰/â€‰mol = 1187

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