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Solved: Consider the balanced equation for the combustion of butane, a fuel often used

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 38P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 38P

PROBLEM 38P

Consider the balanced equation for the combustion of butane, a fuel often used in lighters:

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

Complete the table showing the appropriate masses of reactants and products. If the mass of a reactant is provided, fill in the mass of other reactants required to completely react with the given mass, as well as the mass of each product formed. If the mass of a product is provided, fill in the required masses of each reactant to make that amount of product, as well as the mass of the other product that forms.

Mass CH4H10

Mass O2

Mass CO2

Mass H2O

____

1.1 g

____

____

5.22 g

____

____

____

____

____

10.12 g

____

____

____

____

9.04 g

232 mg

____

____

____

____

____

118 mg

____

Step-by-Step Solution:
Step 1 of 3

Solution 38P

The completed table is as given below

Mass C4H10

Mass O2

Mass CO2

Mass H2O

0.306g

1.1 g

0.930g

0.475g

5.22 g

18.72g

15.84g

8.1g

3.33g

11.96g

10.12 g

5.175g

5.82g

20.89g

17.67g

9.04 g

232 mg

832mg

704mg

360mg

38.8mg

139.4mg

118 mg

60.34mg

Steps for calculation

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

We know from the equation that 2 mole C4H10 combines with 13moles of O2 to give 8 moles  CO2 and 10 moles of H2O

Thus 116 gms of C4H10combines with 416 gms of O2 to give 352 gms  CO2 and 180gms of H2O

1 st row calculation

Mass of C4H10=1.1g=0.306g

Mass of CO2=1.1g=0.930kg

Mass of H2O=1.1g=0.475kg

2nd row calculation

Mass of O2=5.22g=18.72g

Mass of CO2=5.22g=15.84g

Mass of H2O=5.22g=8.1g

3rd row calculation

Mass of C4H10=10.12g=3.33g

Mass of O2=10.12g=11.96g

Mass of CO2=10.12g=5.175 g

4th row calculation

Mass of C4H10=9.04g=5.82g

Mass of O2=9.04g=20.89g

Mass of H2O=9.04g=17.67g

5th row calculation

Mass of O2=232mg=832 mg

Mass of CO2=232mg=704mg

Mass of H2O=232mg=360mg

6th row calculation

Mass of C4H10=118mg=38.8mg

Mass of O2=118mg=139.45mg

Mass of H2O=118mg=60.34mg

Step 2 of 3

Chapter 8, Problem 38P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 38P from 8 chapter was answered, more than 2188 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 38P from chapter: 8 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. This full solution covers the following key subjects: mass, Product, provided, reactant, reactants. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The answer to “Consider the balanced equation for the combustion of butane, a fuel often used in lighters:2 C4H10(g) + 13 O2(g) ? 8 CO2(g) + 10 H2O(g)Complete the table showing the appropriate masses of reactants and products. If the mass of a reactant is provided, fill in the mass of other reactants required to completely react with the given mass, as well as the mass of each product formed. If the mass of a product is provided, fill in the required masses of each reactant to make that amount of product, as well as the mass of the other product that forms.Mass CH4H10Mass O2Mass CO2Mass H2O____1.1 g________5.22 g____________________10.12 g________________9.04 g232 mg____________________118 mg____” is broken down into a number of easy to follow steps, and 110 words.

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Solved: Consider the balanced equation for the combustion of butane, a fuel often used