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Table salt, NaCl(s), and sugar, C1,H„O(s), are accidentally mixed. A 5.50 g sample is burned, and 2.10 g of CO, (g) is produced. What was the mass percentage of the table salt in the mixture? percentage of NaCl:

Table salt, NaCl(s), and sugar, C1,H„O(s), are accidentally mixed. A 5.50 g sample is burned, and 2.10 g of CO, (g) is produced. What was the mass percentage of the table salt in the mixture? percentage of NaCl:

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**General Chemistry 4th Edition** **McQuarrie • Rock • Gallogly** **University Science Books** **Presented by Macmillan Learning** --- Table salt, NaCl(s), and sugar, C₁₂H₂₂O₁₁(s), are accidentally mixed. A 5.50 g sample is burned, and 2.10 g of CO₂(g) is produced. What was the mass percentage of the table salt in the mixture? **percentage of NaCl: [______] %**
Transcribed Image Text:**General Chemistry 4th Edition** **McQuarrie • Rock • Gallogly** **University Science Books** **Presented by Macmillan Learning** --- Table salt, NaCl(s), and sugar, C₁₂H₂₂O₁₁(s), are accidentally mixed. A 5.50 g sample is burned, and 2.10 g of CO₂(g) is produced. What was the mass percentage of the table salt in the mixture? **percentage of NaCl: [______] %**
Expert Solution
Step 1

Balanced chemical equation for combustion of sucrose:

C12H22O11 + 12O2 ------> 12CO2 + 11 H2O 

this equation suggests that:

Number of moles of C12H22O11  = 1/12 Number of moles of CO2

Mass of CO2 produced = 2.10 g

Molar mass of CO2 = 44.01 g/mole

Number of moles of CO2 = mass/molar mass = 2.10/44.01 = 0.04772 mole

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