Chapter 12, Problem 67A

(a)

Interpretation Introduction

Interpretation: The balanced equation for the reaction is to be written.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 67A

The balanced equation for the reaction is as follows:

  ${\text{SF}}_2\left(l\right)+{\text{2F}}_2\left(g\right) \to {\text{SF}}_{\text{6}}\left(g\right)$

Explanation of Solution

Liquid sulfur difluoride reacts with fluorine gas to form gaseous sulfur hexafluoride.

  ${\text{SF}}_2\left(l\right)+{\text{F}}_2\left(g\right) \to {\text{SF}}_{\text{6}}\left(g\right)$

The balanced equation for the reaction is as follows:

  ${\text{SF}}_2\left(l\right)+{\text{2F}}_2\left(g\right) \to {\text{SF}}_{\text{6}}\left(g\right)$

(b)

Interpretation Introduction

Interpretation: The number of fluorine molecules required to react with $5.00\text{mg}$ of sulfur difluoride is to be calculated.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 67A

The number of fluorine molecules required to react with $5.00\text{mg}$ of sulfur difluoride is $8.59\times {10}^{19}\text{molecules}$ .

Explanation of Solution

Mass of sulfur difluoride is $5.00\text{mg}$ .

Molar mass of ${\text{SF}}_{\text{2}}$ is $\text{70}\text{.1}\text{g}$ .

The balanced equation for the reaction is as follows:

  ${\text{SF}}_2\left(l\right)+{\text{2F}}_2\left(g\right) \to {\text{SF}}_{\text{6}}\left(g\right)$

  $\text{1}\text{mol}$ of ${\text{SF}}_{\text{2}}$ reacts with $2\text{mol}$ of ${\text{SF}}_{\text{6}}$ .

The molecules of Fluorine reacted is given as follows:

  $5.00\cancel{\text{mg}{\text{SF}}_2}\times \frac{1\text{}\cancel{\text{g}{\text{SF}}_2}}{{10}^3\cancel{\text{mg}{\text{SF}}_2}}\times \frac{1\cancel{\text{mol}{\text{SF}}_2}}{70.1\cancel{\text{g}{\text{SF}}_2}}\times \frac{2\cancel{\text{mol}{\text{F}}_2}}{1\cancel{\text{mol}{\text{SF}}_2}}\times \frac{6.022\times {10}^{23}{\text{molecule F}}_2}{1\cancel{\text{mol}{\text{F}}_2}}=8.59\times {10}^{19}\text{molecules}$

(c)

Interpretation Introduction

Interpretation: The volume of fluorine gas at STP is to be calculated.

Concept Introduction: Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. A balanced chemical equation is one in which the number of atoms in each of the molecules is the same on both sides of the equation.

Answer to Problem 67A

The volume of fluorine gas at STP is $4.26\text{L}$ .

Explanation of Solution

Mass of sulfur difluoride is $\text{6}\text{.66}\text{g}$ .

Molar mass of ${\text{SF}}_{\text{2}}$ is $\text{70}\text{.1}\text{g}$ .

The balanced equation for the reaction is as follows:

  ${\text{SF}}_2\left(l\right)+{\text{2F}}_2\left(g\right) \to {\text{SF}}_{\text{6}}\left(g\right)$

  $\text{1}\text{mol}$ of ${\text{SF}}_{\text{2}}$ reacts with $2\text{mol}$ of ${\text{SF}}_{\text{6}}$ .

  $\text{1}\text{mol}$ is equal to $\text{22}\text{.4}\text{L}$ at STP.

The volume of fluorine at STP is given as follows:

  $6.66\cancel{\text{mg}{\text{SF}}_2}\times \frac{1\cancel{\text{mol}{\text{SF}}_2}}{70.1\cancel{\text{g}{\text{SF}}_2}}\times \frac{2\cancel{\text{mol}{\text{F}}_2}}{1\cancel{\text{mol}{\text{SF}}_2}}\times \frac{\text{22}\text{.4}{\text{L F}}_2}{1\cancel{\text{mol}{\text{F}}_2}}=4.26\text{L}{\text{F}}_2$