Chapter 4, Problem 28PS

### 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

# At higher temperatures, NaHCO3is converted quantitatively to Na2CO3.2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)Heating a 1.7184-g sample of impure NaHCO3 gives 0.196 g of CO2. What was the mass percent of NaHCO3 in the original 1.7184-g sample?

Interpretation Introduction

Interpretation:

The mass percentage of NaHCO3 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 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 2NaHCO3(s)â€‰â†’Na2CO3(s)â€‰+CO2(g)+H2O(g)

The amount of CO2 was calculated from its mass. Â Because 2 mol of NaHCO3 was present in the sample for each mole of CO2 isolated. Â Therefore amount of NaHCO3 should be known to calculate its mass and mass percentage in the sample.

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

â€‚Â 0.196gâ€‰CO2â€‰Ã—1â€‰molâ€‰CO244.01â€‰gâ€‰CO2=â€‰0.004453â€‰â€‰molâ€‰CO2

1 mol of CO2 is produced from 2 mol of NaHCO3, the amount of NaHCO3 in the sample must also have been 0.004453â€‰â€‰molâ€‰.

Therefore,

â€‚Â 0.004453â€‰â€‰molâ€‰â€‰CO2â€‰Ã—2â€‰molâ€‰NaHCO31â€‰molâ€‰CO2â€‰=â€‰0

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