Chapter 21, Problem 21.148QP

(a)

Interpretation Introduction

Interpretation:

The chemical equation for the given reaction has to be written

Concept Introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

To Write: The chemical equation for the given reaction

Answer to Problem 21.148QP

The chemical equation for the given reaction is: $2\text{Ca}(s)+{\text{ O}}_{\text{2}}(g)\stackrel{}{\to }2\text{CaO}(s)$

Explanation of Solution

Given data:

Burning of calcium metal in oxygen.

Chemical Equation:

The chemical equation for the reaction of burning of calcium metal in oxygen can be written as follows,

$2\text{Ca}(s)+{\text{ O}}_{\text{2}}(g)\stackrel{}{\to }2\text{CaO}(s)$

Conclusion

The chemical equation for the given reaction is written as: $2\text{Ca}(s)+{\text{ O}}_{\text{2}}(g)\stackrel{}{\to }2\text{CaO}(s)$

(b)

Interpretation Introduction

Interpretation:

The chemical equation for the given reaction has to be written

Concept Introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

To Write: The chemical equation for the given reaction

Answer to Problem 21.148QP

The chemical equation for the given reaction is: $4{\text{PH}}_{\text{3}}(g)+{\text{ 8O}}_{\text{2}}(g)\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{10}(S){\text{ + 6H}}_{\text{2}}\text{O}(g)$

Explanation of Solution

Given data:

Phosphine, ${\text{PH}}_{\text{3}}$ burns in excess oxygen

Chemical Equation:

The chemical equation for the reaction of burning of phosphine in excess of oxygen can be written as follows,

$4{\text{PH}}_{\text{3}}(g)+{\text{ 8O}}_{\text{2}}(g)\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{10}(S){\text{ + 6H}}_{\text{2}}\text{O}(g)$

Conclusion

The chemical equation for the given reaction is written as: $4{\text{PH}}_{\text{3}}(g)+{\text{ 8O}}_{\text{2}}(g)\stackrel{}{\to }{\text{P}}_{\text{4}}{\text{O}}_{10}(S){\text{ + 6H}}_{\text{2}}\text{O}(g)$

(c)

Interpretation Introduction

Interpretation:

The chemical equation for the given reaction has to be written

Concept Introduction:

Balanced Chemical equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

To Write: The chemical equation for the given reaction

Answer to Problem 21.148QP

The chemical equation for the given reaction is: $4{\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}(l)+{\text{ 13O}}_{\text{2}}(g)\stackrel{}{\to }{\text{8CO}}_{\text{2}}(g){\text{ + 14H}}_{\text{2}}\text{O}(g){\text{ + 2N}}_{\text{2}}(g)$

Explanation of Solution

Given data:

Ethanolamine, ${\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$ burns in excess of oxygen.

The $\text{N}$ atom of ethanolamine ends up as ${\text{N}}_{\text{2}}$

Chemical Equation:

The chemical equation for the reaction of burning of ethanolamine in excess oxygen by giving out nitrogen gas as one of the product can be written as follows,

$4{\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}(l)+{\text{ 13O}}_{\text{2}}(g)\stackrel{}{\to }{\text{8CO}}_{\text{2}}(g){\text{ + 14H}}_{\text{2}}\text{O}(g){\text{ + 2N}}_{\text{2}}(g)$

Conclusion

The chemical equation for the given reaction is written as: $4{\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}(l)+{\text{ 13O}}_{\text{2}}(g)\stackrel{}{\to }{\text{8CO}}_{\text{2}}(g){\text{ + 14H}}_{\text{2}}\text{O}(g){\text{ + 2N}}_{\text{2}}(g)$