Chapter 3, Problem 65QRT

(a)

Interpretation Introduction

Interpretation:

The balanced equation for ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ reacts with sodium oxalate gives carbon, carbon dioxide, sodium chloride, and sodium fluoride has to be written.

Concept introduction:

Balancing reaction:

Balanced reaction is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the information

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.
• The coefficient should be reduced to the smallest whole number.

Explanation of Solution

The given statement is shown below,

   ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ reacts with sodium oxalate gives carbon, carbon dioxide, sodium chloride, and sodium fluoride.

Balance the equation,

    ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + CO}}_{\text{2}}\text{ + NaCl + NaF}$

Balance the equation, when balancing the equation, we should not alter the subscripts and we can change coefficients. There are two fluorine atoms in the left side and one fluorine atoms in the right side. Therefore, two molecule of sodium fluoride are added to right side of reaction. Now, the balanced equation is given below.

  ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + CO}}_{\text{2}}\text{ + NaCl + 2NaF}$

There are two chlorine atoms in the left side and one chlorine atoms in the right side. Therefore, two molecule of sodium chloride are added to right side of reaction. Now, the balanced equation is given below.

    ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + CO}}_{\text{2}}\text{ + 2NaCl + 2NaF}$

There are two sodium atoms in the left side and four sodium atoms in the right side. Therefore, two molecule of sodium oxalate are added to left side of reaction. Now, the balanced equation is given below.

    ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + 2 Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + CO}}_{\text{2}}\text{ + 2NaCl + 2NaF}$

There are eight oxygen atoms in the left side and two oxygen atoms in the right side. Therefore, four carbon dioxide molecules are added to right side of reaction. Now, the balanced equation is given below.

  ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + 2 Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + 4 CO}}_{\text{2}}\text{ + 2 NaCl + 2 NaF}$

(b)

Interpretation Introduction

Interpretation:

Mass of sodium oxalate required to remove $\text{76}\text{.8 g}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ has to be calculated.

Concept introduction:

Number of moles can be calculated by using following formula,

  $\text{No}\text{.of Moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

The reaction is shown below,

  ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + 2 Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + 4 CO}}_{\text{2}}\text{ + 2 NaCl + 2 NaF}$

Molar Mass ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ is 133}\text{.9986 g/mol}$

Given,

  $\text{76}\text{.8 g}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$

  $\begin{array}{l}\text{No}\text{.of Moles = }\frac{\text{Mass}\text{of}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}\\ \text{No}\text{.of Moles = }\frac{76.8\text{g}}{\text{120}\text{.914 }\text{g/mol}}\\ \text{No}\text{.of Moles =}\text{0}\text{.635}\text{mol}\end{array}$

According to the balanced equation, 2 moles of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ reaction with 1 moles of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$. Therefore,

  $\begin{array}{l}\text{No}{\text{.of Moles Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ = 0}\text{.635}\text{mol}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}\times \frac{2{\text{mol Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{1 mol}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}\\ \text{No}{\text{.of Moles Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ =}\text{1}\text{.27}{\text{mol Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\end{array}$

Grams (mass) of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is calculated as follows,

    $\begin{array}{l}\text{Mass}{\text{of Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ = Moles}\text{×Molar}\text{mass}\\ \text{Mass}{\text{of Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ = 1}\text{.27}{\text{molNa}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{×}\frac{133.9986{\text{g of Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{1mol}\text{of}{\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}\\ \text{Mass}{\text{of Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\text{ =}170.22{\text{g Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\end{array}$

Mass in grams of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$ is $170.22\text{g}$.

(c)

Interpretation Introduction

Interpretation:

Mass in grams of ${\text{CO}}_2$ has to be calculated.

Concept introduction:

Refer to part (b)

Explanation of Solution

The reaction is shown below,

  ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}{\text{ + 2 Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\to {\text{C + 4 CO}}_{\text{2}}\text{ + 2 NaCl + 2 NaF}$

Molar Mass ${\text{CO}}_2\text{ is 44}\text{.0095 g/mol}$

Given,

  $\text{76}\text{.8 g}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$

  $\begin{array}{l}\text{No}\text{.of Moles = }\frac{\text{Mass}\text{of}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}\\ \text{No}\text{.of Moles = }\frac{76.8\text{g}}{\text{120}\text{.914 }\text{g/mol}}\\ \text{No}\text{.of Moles =}\text{0}\text{.635}\text{mol}\end{array}$

According to the balanced equation, 1 moles of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ gives 4 moles of ${\text{CO}}_2$. Therefore,

  $\begin{array}{l}\text{No}{\text{.of Moles CO}}_2\text{ = 0}\text{.635}\text{mol}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}\times \frac{4{\text{mol CO}}_2}{\text{1 mol}{\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}}\\ \text{No}{\text{.of Moles CO}}_2\text{ =}\text{2}\text{.54}{\text{mol CO}}_2\end{array}$

Grams (mass) of ${\text{CO}}_2$ is calculated as follows,

    $\begin{array}{l}\text{Mass}{\text{of CO}}_2\text{ = Moles}\text{×Molar}\text{mass}\\ \text{Mass}{\text{of CO}}_2\text{ = 2}\text{.54}\text{mol}{\text{CO}}_2\text{×}\frac{44.0095{\text{g of CO}}_2}{\text{1mol}\text{of}{\text{CO}}_2}\\ \text{Mass}{\text{of CO}}_2\text{ =}111.78{\text{g CO}}_2\end{array}$

Mass in grams of ${\text{CO}}_2$ is $111.78\text{g}$.