   # At higher temperatures, NaHCO 3 is converted quantitatively to Na 2 CO 3 . 2 NaHCO 3 (s) → Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g) Heating a 1.7184-g sample of impure NaHCO 3 gives 0.196 g of CO 2 . What was the mass percent of NaHCO 3 in the original 1.7184-g sample? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 28PS
Textbook Problem
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## 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 of Solution

Balanced chemical equation for the given reaction is,

2NaHCO3(s)Na2CO3(s)+CO2(g)+H2O(g)

The amount of CO2 was calculated from its mass.  Because 2 mol of NaHCO3 was present in the sample for each mole of CO2 isolated.  Therefore amount of NaHCO3 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 Carbon dioxide is,

0.196gCO2×1molCO244.01gCO2=0.004453molCO2

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

Therefore,

0.004453molCO2×2molNaHCO31molCO2=0

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