Chapter 4.9, Problem 4.3ACP

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calculate the volume of water obtained from 1.00 liter of liquid oxygen. (a) 2.0 L (b) 1.28 L (c) 0.64 L (d) 1.0 L

Interpretation Introduction

Interpretation:

The volume of water obtained from 1.00L of liquid oxygen has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

Numberofmole=MassingramMolarmass

Massingramofthesubstance=Numberofmole×Molarmass

• 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 mole (g/mol).
• Mass=Density×Volume
• Volume=MassDensity
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
Explanation

Reason for the correct option:

The strategy map is,

Â  Volâ€‰O2â€‰â†’1â€‰Aâ€‰â†’2â€‰molâ€‰O2â€‰â†’3â€‰â€‰Bâ€‰â€‰â†’4Massâ€‰â€‰ofâ€‰H2Oâ†’5â€‰â€‰Volâ€‰H2O

Here A is the mass of O2 and B is the amount of H2O.

The balanced chemical equation for the reaction involving hydrogen and oxygen as a reactant is,

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  2H2+â€‰O2â€‰â†’â€‰â€‰2â€‰H2O

The amount of O2 available can be calculated as follows,

Â Â  Â Â Â Â  Â Â 1.00â€‰Ã—103â€‰mLâ€‰O2â€‰Ã—(â€‰1.14â€‹â€‰gâ€‰O21â€‰mLâ€‰O2)â€‰Ã—(1â€‰molâ€‰O232.00â€‰â€‰gâ€‰O2)â€‰=â€‰35.625â€‰mol Â Â Â Â Â Â Â Â Â Â Â

From the amount of O2 number of moles of H2O(B) can be calculated. Stoichiometric ratio between O2 and H2O is 2:1

So,

Â Â Â Â Â  The amount ofÂ  H2O =Â  35.625â€‰molâ€‰O2â€‰Ã—2â€‰molâ€‰H2O1â€‰molâ€‰O2â€‰=â€‰71

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started