Chapter 6, Problem 140QP

(a)

Interpretation Introduction

Interpretation:

The decomposition of 1.0 mol of the given compound produces 1.5 mol of ${\text{N}}_2$ or not is to be identified.

Explanation of Solution

4 moles of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ decomposes to produce 6 moles of ${\text{N}}_2$ . The number of moles of ${\text{N}}_2$ produced by 1.0 mol of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\begin{array}{c}4{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=6{\text{ mol N}}_{\text{2}}\\ 1.0{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=\frac{6{\text{ mol N}}_{\text{2}}}{4\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}}\times 1.0\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}\\ =1.5{\text{ mol N}}_{\text{2}}\end{array}$

Hence, option A is correct.

(b)

Interpretation Introduction

Interpretation:

The decomposition of 1.0 mol of the given compound produces 3.0 mol of ${\text{CO}}_2$ or not is to be identified.

Explanation of Solution

4 moles of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ decomposes to produce 12 moles of ${\text{CO}}_2$ .

The number of moles of ${\text{CO}}_2$ produced by 1.0 mol of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\begin{array}{c}4{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=12{\text{ mol CO}}_2\\ 1.0{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=\frac{12{\text{ mol CO}}_2}{4\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}}\times 1.0\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}\\ =3.0{\text{ mol CO}}_2\end{array}$

Hence, option B is correct.

(c)

Interpretation Introduction

Interpretation:

The decomposition of 1.0 mol of the given compound produces 2.5 moles of ${\text{H}}_2\text{O}$ or not is to be identified.

Explanation of Solution

4 moles of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ react to produce 10 moles of ${\text{H}}_2\text{O}$ .

The number of moles of ${\text{H}}_2\text{O}$ produced by 1.0 mol of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\begin{array}{c}4{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=10{\text{ mol H}}_2\text{O}\\ 1.0{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=\frac{10{\text{ mol H}}_2\text{O}}{4\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}}\times 1.0\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}\\ =2.5{\text{ mol H}}_2\text{O}\end{array}$

Hence, option C is correct.

(d)

Interpretation Introduction

Interpretation:

The decomposition of 1.0 mol of the given compound produces 1.0 mole of ${\text{O}}_2$ or not is to be identified.

Explanation of Solution

4 moles of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ react to produce 1 mole of ${\text{O}}_2$ .

The number of moles of ${\text{O}}_2$ produced by 1.0 mol of ${\text{C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9$ is calculated as follows:

$\begin{array}{c}4{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=1{\text{ mol O}}_2\\ 1.0{\text{ mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9=\frac{1{\text{ mol O}}_2}{4\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}}\times 1.0\sout{{\text{mol C}}_3{\text{H}}_3{\text{N}}_3{\text{O}}_9}\\ =0.25{\text{ mol O}}_2\end{array}$

Hence, option D is incorrect and the correct value of the product is 0.25 mol of ${\text{O}}_2$ .

(e)

Interpretation Introduction

Interpretation:

Whether all the above options (a,b,c,d) are correct or not is to be identified.

Explanation of Solution

Among the first four options, the (d) part is incorrect. Thus the given statement that“ all the options are correct� is proved to be a wrong statement.