Chapter 9, Problem 10A

(a)

Interpretation Introduction

Interpretation:

Moles of oxygen must be found out when 0.625 moles of KOH is produced from the reaction of KO₂ and H₂O.

Concept Introduction:

Four mole KO₂ reacts with two moles of H₂O to produce three moles of O₂ and four moles of KOH.

Answer to Problem 10A

0.469 mole of O₂ will be produced.

Explanation of Solution

  ${\text{KO}}_{\text{2}}{\text{(s)+2H}}_{\text{2}}\text{O(l)}\to {\text{3O}}_{\text{2}}\text{(g)+4KOH(s)}\text{.}$

From the balanced equation it is clear that when 4 moles of KOH is produced then 3 moles of O₂ is produced.

So, when 1 mole of KOH is produced then $\frac{3}{4}$ moles of O₂ is produced.

Thus, when 0.625 mole of KOH is produced then $\frac{\text{3×0}\text{.625}}{\text{4}}\text{=0}\text{.4687}\approx \text{0}\text{.469}$ moles of O₂ is produced.

b)

Interpretation Introduction

Interpretation:

Moles of selenium must be found out when 0.625 moles of H₂O is produced from the reaction of SeO₂ and H₂Se.

Concept Introduction:

One mole SeO₂ reacts with two moles of H₂Se to produce three moles of Se and two moles of H₂O.

Answer to Problem 10A

0.938 mole of Se will be produced.

Explanation of Solution

  ${\text{SeO}}_{\text{2}}{\text{(g)+2H}}_{\text{2}}\text{Se(g)}\to {\text{3Se(s)+2H}}_{\text{2}}\text{O(g)}\text{.}$

From the balanced equation it is clear that when 2 mole of H₂O is produced then 3 moles of Se is produced.

So, when 1 mole of H₂O is produced then $\frac{3}{2}$ moles of Se is produced.

Thus, when 0.625 mole of H₂O is produced then $\frac{\text{3×0}\text{.625}}{\text{2}}\text{=0}\text{.9375}\approx \text{0}\text{.938}$ moles of Se is produced.

c)

Interpretation Introduction

Interpretation:

Moles of acetaldehyde must be found out when 0.625 moles of H₂O is produced from the reaction of CH₃CH₂OH and O₂.

Concept Introduction:

One mole CH₃CH₂OH reacts with 0.5 mole of O₂ to produce one moles of CH₃CHO and one mole of H₂O.

Answer to Problem 10A

0.625 mole of CH₃CHO will be produced.

Explanation of Solution

  ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(l)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{CH}}_{\text{3}}{\text{CHO(aq)+H}}_{\text{2}}\text{O(l)}\text{.}$

From the balanced equation it is clear that when 1 mole of H₂O is produced then 1 mole of CH₃CHO is produced.

So, when 0.625 mole of H₂O is produced then 0.625 moles of CH₃CHO is produced.

d)

Interpretation Introduction

Interpretation:

Moles of iron must be found out when 0.625 moles of Al₂O₃ is produced from the reaction of Fe₂O₃ and Al.

Concept Introduction:

One mole Fe₂O₃ reacts with 2 moles of Al to produce one moles of Al₂O₃ and two moles of Fe.

Answer to Problem 10A

1.25 mole of Fe will be produced.

Explanation of Solution

  ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)+2Al(s)}\to {\text{2Fe(s)+Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{.}$

From the balanced equation it is clear that when 1 mole of Al₂O₃ is produced then 2 moles of Fe is produced.

So, when 0.625 mole of Al₂O₃ is produced then $0.625\times 2=1.25$ moles of Fe is produced.