Chapter 4.9, Problem 4.4ACP

### 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

# For Further Thought: The density of hydrazine, N2H4, is 1.02 g/mL and its molar mass is 32.05 g/mol. If all of the hydrogen in 1.0 L of hydrazine is converted to water, what volume of water will be formed? Look back at your earlier prediction. The analogy between liquid volume and atom count is quite poor. Speculate on why this estimate was not very good.

Interpretation Introduction

Interpretation:

The volume of water obtained from 1.00L of hydrazine has to be determined. And it has to be explained that why this calculation is not very good as the calculation of volume of water from liquid oxygen.

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

The balanced chemical equation for the reaction involving hydrazine 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  N2H4+â€‰O2â€‰â†’â€‰N2â€‰+â€‰2â€‰H2O

The amount of N2H4 available can be calculated as follows,

Â Â  Â Â Â Â Â Â  1.00â€‰Ã—103â€‰mLâ€‰N2H4â€‰Ã—(â€‰1.02gâ€‰N2H41â€‰mLâ€‰N2H4)â€‰Ã—(1â€‰molâ€‰N2H432.00â€‰â€‰gâ€‰N2H4)â€‰=â€‰31.875â€‰mol Â Â Â Â Â Â Â Â Â Â Â

From the amount of N2H4 number of moles of H2O can be calculated. Stoichiometric ratio between N2H4 and H2O is 2:1

So,

Â Â Â Â Â  The amount ofÂ  H2O =Â  31.875â€‰molâ€‰N2H4â€‰Ã—2â€‰molâ€‰H2O1â€‰molâ€‰N2H4â€‰=â€‰63

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