Chapter 11.2, Problem 14PP

(a)

Interpretation Introduction

Interpretation:

The mass of chlorine gas that reacts with given moles of titanium oxide are to be interpreted.

Concept introduction:

The formula to calculate the number of moles is given as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 14PP

The mass of chlorine required is $177.5\text{g}$.

Explanation of Solution

The given chemical equation is shown below:

${\text{TiO}}_{\text{2}}\left(\text{s}\right)+\text{C}\left(\text{s}\right)+2{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{TiCl}}_4\left(\text{s}\right)+{\text{CO}}_2\left(\text{g}\right)$

The number of moles of chlorine that reacts with given moles of titanium oxide are calculated as shown below:

$\begin{array}{c}\text{1}\text{mol}{\text{TiO}}_{\text{2}}=2\text{mol}{\text{Cl}}_{\text{2}}\\ \text{1}\text{.25}\text{mol}{\text{TiO}}_{\text{2}}=2\times 1.25\text{mol}{\text{Cl}}_{\text{2}}\\ =2.5\text{mol}{\text{Cl}}_{\text{2}}\end{array}$

The molar mass of chlorine gas is $\text{71}\text{.0}\text{g/mol}$. Substitute the number of moles and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{2}\text{.5}\text{mol}=\frac{\text{Given mass}}{\text{71}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=\text{2}\text{.5}\text{mol}\times \text{71}\text{.0}\frac{\text{g}}{\text{mol}}\\ =177.5\text{g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The mass of carbon that reacts with given moles of titanium oxide are to be interpreted.

Concept introduction:

The formula to calculate the number of moles is given as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 14PP

The mass of carbon is $15.0\text{g}$.

Explanation of Solution

The given chemical equation is shown below:

${\text{TiO}}_{\text{2}}\left(\text{s}\right)+\text{C}\left(\text{s}\right)+2{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to {\text{TiCl}}_4\left(\text{s}\right)+{\text{CO}}_2\left(\text{g}\right)$

The number of moles of carbon that reacts with given moles of titanium oxide are calculated as shown below:

$\begin{array}{c}\text{1}\text{mol}{\text{TiO}}_{\text{2}}=1\text{mol}\text{C}\\ \text{1}\text{.25}\text{mol}{\text{TiO}}_{\text{2}}=1\times 1.25\text{mol}\text{C}\\ =1.25\text{mol}\text{C}\end{array}$

The molar mass of carbon is $\text{12}\text{.0}\text{g/mol}$. Substitute the number of moles and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{1}\text{.25}\text{mol}=\frac{\text{Given mass}}{\text{12}\text{.0}\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1.25\text{mol}\times 12.0\frac{\text{g}}{\text{mol}}\\ =15.0\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The mass of products formed from given moles of titanium oxide are to be interpreted.

Concept introduction:

The formula to calculate the number of moles is given as shown below:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 14PP

The mass of products produced is $\text{292}\text{.3}\text{g}$.

Explanation of Solution

The molar mass of titanium oxide is $\text{79}\text{.866}\text{g/mol}$. Substitute the number of moles and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{1}\text{.25}\text{mol}=\frac{\text{Given mass}}{79.866\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=1.25\text{mol}\times 79.866\frac{\text{g}}{\text{mol}}\\ =99.8\text{g}\end{array}$

Total mass of reactants is calculated as shown below:

$\begin{array}{c}\text{Total mass of reactants}={\text{Mass of Cl}}_{\text{2}}+{\text{Mass of TiO}}_{\text{2}}+\text{Mass of C}\\ \text{=177}\text{.5}\text{g}+\text{99}\text{.8}\text{g}+\text{15}\text{.0}\text{g}\\ =\text{292}\text{.3}\text{g}\end{array}$

According to law of conservation of mass, the mass of reactants and products is equal to each other. Thus, the mass of products produced is also $\text{292}\text{.3}\text{g}$.