Chapter 9, Problem 10A

(a)

Interpretation Introduction

Interpretation:

The number of moles of the first product produced needs to be determined if 0.625 mole of the second product forms.

  $K{O}_{2\left(s\right)}+H_2{O}_{\left(l\right)}\to O_{2\left(g\right)}+KO{H}_{\left(s\right)}$

Concept Introduction:

To calculate moles of first product formed, balance the equation representing the reaction then use mol ratios to calculate the unknown moles of the product.

The balanced equation is:

  $4K{O}_{2\left(s\right)}+2H_2{O}_{\left(l\right)}\to 3O_{2\left(g\right)}+4KO{H}_{\left(s\right)}$

Answer to Problem 10A

0.469 mole of first product is formed.

Explanation of Solution

The balanced equation is:

  $4K{O}_{2\left(s\right)}+2H_2{O}_{\left(l\right)}\to 3O_{2\left(g\right)}+4KO{H}_{\left(s\right)}$

From the equation above, first product is O₂ while the second product is KOH.

Hence,

The mol ratio of first product to second product is 3:4.

If 0.625 mole of the second product is formed, then number of moles of first product formed is:

  $\frac{3}{4}\times 0.625mole=0.469mole$

0.469 mole of first product is formed.

(b)

Interpretation Introduction

Interpretation:

The number of moles of the first product produced if 0.625 mole of the second product forms needs to be determined.

  $Se{O}_{2\left(g\right)}+H_{2}S{e}_{\left(g\right)}\to S{e}_{\left(s\right)}+H_2{O}_{\left(l\right)}$

Concept Introduction:

To calculate moles of first product formed, balance the equation representing the reaction then use mol ratios to calculate the unknown moles of the product.

The balanced equation is:

  $Se{O}_{2\left(g\right)}+2H_{2}S{e}_{\left(g\right)}\to 3S{e}_{\left(s\right)}+2H_2{O}_{\left(l\right)}$

Answer to Problem 10A

0.938 moleof first product is formed.

Explanation of Solution

The balanced equation is:

  $Se{O}_{2\left(g\right)}+2H_{2}S{e}_{\left(g\right)}\to 3S{e}_{\left(s\right)}+2H_2{O}_{\left(l\right)}$

From the reaction above, the first product is Se while the second product is H₂O.

Hence, mol ratio of Se:H₂O is 3:2. Therefore, number of moles of the first product formed is:

  $\frac{3}{2}\times 0.625mole=0.938mole$

(c)

Interpretation Introduction

Interpretation:

The number of moles of the first product produced if 0.625 mole of the second product forms needs to be determined.

  $C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}+O_{2\left(g\right)}\to C{H}_{3}CH{O}_{\left(aq\right)}+H_2{O}_{\left(l\right)}$

Concept Introduction:

To calculate moles of first product formed, balance the equation representing the reaction then use mol ratios to calculate the unknown moles of the product.

The balanced equation is:

  $2C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}+O_{2\left(g\right)}\to 2C{H}_{3}CH{O}_{\left(aq\right)}+2H_2{O}_{\left(l\right)}$

Answer to Problem 10A

c) 0.625 moleof first product is formed.

Explanation of Solution

The balanced equation is:

  $2C{H}_{3}C{H}_{2}O{H}_{\left(l\right)}+O_{2\left(g\right)}\to 2C{H}_{3}CH{O}_{\left(aq\right)}+2H_2{O}_{\left(l\right)}$

From the equation, the first product is CH₃CHO and the second product is H₂O.

Mol ratio of the first to the second product is 2:2, which can be simplified into 1:1.

Number of moles of CH₃CHO produced equals number of moles of H₂O_.

Hence, number of moles of first product formed is 0.625 mole.

(d)

Interpretation Introduction

Interpretation:

The number of moles of the first product produced if 0.625 mole of the second product forms needs to be determined.

  $\begin{array}{l} \hfill \\ F{e}_2{O}_{3\left(s\right)}+A{l}_{\left(s\right)}\to F{e}_{\left(l\right)}+A{l}_2{O}_{3\left(s\right)}\hfill \\ \hfill \end{array}$

Concept Introduction:

To calculate moles of first product formed, balance the equation representing the reaction then use mol ratios to calculate the unknown moles of the product.

The balanced equation is:

  $\begin{array}{l} \hfill \\ F{e}_2{O}_{3\left(s\right)}+2A{l}_{\left(s\right)}\to 2F{e}_{\left(l\right)}+A{l}_2{O}_{3\left(s\right)}\hfill \\ \hfill \end{array}$

Answer to Problem 10A

1.250 moleof first product is formed.

Explanation of Solution

The balanced equation is:

  $\begin{array}{l} \hfill \\ F{e}_2{O}_{3\left(s\right)}+2A{l}_{\left(s\right)}\to 2F{e}_{\left(l\right)}+A{l}_2{O}_{3\left(s\right)}\hfill \\ \hfill \end{array}$

From the reaction, Fe is the first product whereas Al₂O₃ is the second product.

Mol ratio of Fe:Al₂O₃ is 2:1.

Hence,

Given that 0.625 mole of Al₂O₃ is formed, number of moles of Fe formed is:

  $\frac{2}{1}\times 0.625mole=1.250mole$