Chapter 13, Problem 93A

(a)

Interpretation Introduction

Interpretation: Amount of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ in g is to be determined.

Concept introduction:

Expression for the calculation of mole is as follows:

  $\text{mole}=\frac{\text{mass in g}}{\text{molar mass}}$

Answer to Problem 93A

  $\text{51}\text{.1 g}$

Explanation of Solution

Given reaction is as follows:

  ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}+{\text{H}}_{\text{2}}\text{O}\to 2{\text{HClO}}_{\text{4}}$

Amount of ${\text{HClO}}_{\text{4}}$ is $56.2\text{ g}$ .

Molecular mass of $2{\text{HClO}}_{\text{4}}$ is $100.46 \text{g/mol}$ .

Calculate the no. of mole of $2{\text{HClO}}_{\text{4}}$ as shown,

  $\begin{array}{c}\text{mole}=\frac{\text{mass in g}}{\text{molar mass}}\\ =\frac{56.2\text{ g}}{100.46 \text{g/mol}}\\ =\text{0}\text{.559 mol}\end{array}$

From the above reaction, 1 mole of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ is required for 2 moles of $2{\text{HClO}}_{\text{4}}$ .

So, calculate moles of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ as follows:

  $\begin{array}{c}\text{no}{\text{. of moles of Cl}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{{\text{1 mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}}{2\text{ mol }\bcancel{{\text{HClO}}_{\text{4}}}}\times \text{0}\text{.559 mol }\bcancel{{\text{HClO}}_{\text{4}}}\\ =\text{0}{\text{.2795 mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}\end{array}$

Molar mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ is $182.901 \text{ g/mol}$ .

Now, calculate mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ as follows:

  $\begin{array}{c}\text{mass in g}=\text{mass}\times \text{molar mass}\\ =\text{0}\text{.2795 g}\times 182.901 \text{ g/mol}\\ =\text{51}\text{.1 g}\end{array}$

Thus, the mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ required is $\text{51}\text{.1 g}$ .

(b)

Interpretation Introduction

Interpretation: Volume of water is to be determined.

Concept introduction:

Expression for the calculation of mole is as follows:

  $\text{mole}=\frac{\text{volume in mL}}{\text{volume of 1 mol}\left(22400\text{ mL}\right)}$

Answer to Problem 93A

  $\text{38,080 mL}$

Explanation of Solution

The given reaction is as follows:

  ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}+{\text{H}}_{\text{2}}\text{O}\to 2{\text{HClO}}_{\text{4}}$

Amount of ${\text{HClO}}_{\text{4}}$ is $3.40\text{ mol}$ .

From the above reaction, 1 mole of ${\text{H}}_{\text{2}}\text{O}$ is required for 2 moles of $2{\text{HClO}}_{\text{4}}$ .

So, calculate moles of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ as follows:

  $\begin{array}{c}\text{no}{\text{. of moles of H}}_{\text{2}}\text{O}=\frac{{\text{1 mol H}}_{\text{2}}\text{O}}{2\text{ mol }\bcancel{{\text{HClO}}_{\text{4}}}}\times \text{3}\text{.40 mol }\bcancel{{\text{HClO}}_{\text{4}}}\\ =\text{1}\text{.7 mol}\end{array}$

Volume of 1 mol is 22400 mL.

Now, calculate the volume of water as follows:

  $\begin{array}{c}\text{volume in mL}=\text{1}\text{.7 mol}\times 22400\text{ mL}\\ =\text{38,080 mL}\end{array}$

Thus, the volume of water required is $\text{38,080 mL}$ .