Explanation Given, 7 .290-mg mixture containing only cyclohexane and acetaldehyde, which yielded 21 .999 mg CO 2 by combustion reaction. The balanced equation is shown below, C 6 H 12 (l) + O 2 (g) → CO 2 (g) + H 2 O(l) While balancing the equation, the subscripts cannot be altered but coefficients can be changed. There are six carbon atoms in the left side of the reaction and one carbon atom in the right side of the reaction, therefore six molecule of CO 2 is added to the right side of the reaction Now, the balanced equation is given below. C 6 H 12 (l) + O 2 (g) → 6 CO 2 (g) + H 2 O(l) There are twelve hydrogen atoms in the left side of the reaction and two hydrogen atoms in the right side of the reaction, therefore six molecule of H 2 O is added to the right side of the reaction. Now, the balanced equation is given below. C 6 H 12 (l) + O 2 (g) → 6 CO 2 (g) + 6H 2 O(l) There are two oxygen atoms in the left side of the reaction and eighteen oxygen atoms in the right side of the reaction, therefore nine molecule of O 2 is added to the left side of the reaction. Now, the balanced equation is given below. C 6 H 12 (l) + 9O 2 (g) → 6 CO 2 (g) + 6H 2 O(l) Similarly balance the same combustion equation using acetaldehyde, C 2 H 4 O(l) + O 2 (g) → CO 2 (g) + H 2 O(l) While balancing the equation, the subscripts cannot be altered but coefficients can be changed. There are two carbon atoms in the left side of the reaction and one carbon atom in the right side of the reaction, therefore two molecule of CO 2 is added to the right side of the reaction Now, the balanced equation is given below. C 2 H 4 O(l) + O 2 (g) → 2 CO 2 (g) + H 2 O(l) There are two hydrogen atoms in the right side of the reaction and four hydrogen atoms in the left side of the reaction, therefore two molecule of H 2 O is added to the right side of the reaction. Now, the balanced equation is given below. C 2 H 4 O(l) + O 2 (g) → 2 CO 2 (g) + 2H 2 O(l) There are six oxygen atoms in the right side of the reaction and three oxygen atoms in the left side of the reaction, therefore 5 2 molecule of O 2 is added to the right side of the reaction. Now, the balanced equation is given below. C 2 H 4 O(l) + 5 2 O 2 (g) → 2 CO 2 (g) + 2H 2 O(l) Let consider, X is the mass of milligrams of acetaldehyde and Y is the mass of milligrams of cyclohexane. Therefore, the total mass of the sample can be given as shown below. X + Y= 7 .290 mg Y= 7 .290 mg - X Molar Mass of CO 2 is 44 .0095 g/mol , Molar Mass of C 2 H 4 O is 44 .0095 g/mol and Molar Mass of C 6 H 12 is 84 .1590 g/mol The number of moles CO 2 is calculated as follows

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

5th Edition

ISBN: 9781285460420

Author: John W. Moore; Conrad L. Stanitski

Publisher: Cengage Learning US

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Chapter 3, Problem 152QRT

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The mass percent of acetaldehyde in the mixture has to be calculated.

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Chapter 3, Problem 151QRT

Chapter 3, Problem 153QRT

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OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

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The mass percent of acetaldehyde in the mixture has to be calculated.

Solution Summary: The author calculates the mass percent of acetaldehyde in the mixture by balancing the equation.

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Interpretation: The mass percent of acetaldehyde in the mixture has to be calculated.

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Chapter 3 Solutions

Interpretation:

The mass percent of acetaldehyde in the mixture has to be calculated.