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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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.