Chapter 11, Problem 66A

(a)

Interpretation Introduction

Interpretation:

The balance chemical equation for the reaction of liquid hydrazine with liquid hydrogen peroxide needs to be determined.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 66A

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\stackrel{}{\to }{\text{ N}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

The reaction of solution of reaction of liquid hydrazine with liquid hydrogen peroxide produces a nitrogen gas with liquid water. It is used to fuel rockets.

The balance reaction must have all the atoms same at both sides of the reaction equation. Thus the balance equation can be written as:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\stackrel{}{\to }{\text{ N}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(b)

Interpretation Introduction

Interpretation:

The mass of hydrazine required to produce 10.0 moles of nitrogen gas in the given reaction needs to be determined.

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\stackrel{}{\to }{\text{ N}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can write as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 66A

${\text{320 g N}}_{\text{2}}{\text{H}}_{\text{4}}$

Explanation of Solution

The reaction of solution of reaction of liquid hydrazine with liquid hydrogen peroxide produces a nitrogen gas with liquid water.

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(l)}}{\text{ + 2 H}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(l)}}\stackrel{}{\to }{\text{ N}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The mole ratio relates the moles of reactant with other reactant and product molecules in the balance chemical equation.

Moles of nitrogen = 10.0 moles

Molar mass of hydrazine = 32 g/mol

1 mole of nitrogen = 1 mole of hydrazine

Calculate mass of nitrogen with the help of mole ratio:

$\text{10}{\text{.0 moles N}}_{\text{2}}\text{ ×}\frac{{\text{1 mole N}}_{\text{2}}{\text{H}}_{\text{4}}}{{\text{1 moles N}}_{\text{2}}}\text{×}\frac{{\text{ 32 g N}}_{\text{2}}{\text{H}}_{\text{4}}}{{\text{1 mole N}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{= 320 g N}}_{\text{2}}{\text{H}}_{\text{4}}$