Q4.98P - Chemistry: Molecular Nature Of Matter And Change

Question

How many grams of chlorine gas can be produced from the electrolytic decomposition of 874 g of calcium chloride? Name and calculate the mass (in g) of the other product.

Step-by-Step Solution

Answer

You need to calculate how many grams of chlorine gas can be prepared from the electrolytic decomposition of 874g of calcium chloride. Also you need to write the name and calculate the mass (in g) of the other product.

Step 1: Balanced equation

Electrolytic decomposition of calcium chloride forms calcium and chlorine gas. The balanced equation is

 $\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\stackrel{}{\mathbf{\to }}\mathit{C}\mathit{a}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

 $\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\stackrel{}{\mathbf{\to }}\mathit{C}\mathit{a}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step 2: Calculation of moles of CaCl 2

According to the question,

 

Mass of CaCl₂ = 874g

 

Molecular mass of CaCl₂ = 110g/mol

 

Again you know,

 

$\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{\left(}\mathbf{g}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{\right)}}{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{\right)}}$ 

Moles of CaCl₂

$\begin{array}{l}\mathbf{=}\frac{\mathbf{874}}{\mathbf{110}}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\\ \mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\end{array}$ 

 

Hence, moles of CaCl₂ are 7.9454mol.

Step 3: Calculation of moles of Cl 2

According to the balanced equation

 $\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\stackrel{}{\mathbf{\to }}\mathit{C}\mathit{a}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

 

1 mole of CaCl₂ produces 1 mole Cl₂

 

Now, moles of Cl₂

 $\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{1}}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\\ \mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\end{array}$

 

Hence, moles of Cl₂ are 7.9454mol.

Step 4: Calculation of mass of Cl 2

Again you know,

 $\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{\times }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{\right)}$

 

Molar mass of Cl₂ is 70g/mol.

 

Now, mass of Cl₂

$\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\times }\mathbf{70}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{=}\mathbf{556}\mathbf{.178}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\end{array}$ 

 

Hence, 556.178g of Cl₂ gas can be prepared from the electrolytic decomposition of 874g of CaCl₂.

 

 

Step 5: Calculation of moles of Ca

According to the balanced equation

 

$\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\stackrel{}{\mathbf{\to }}\mathit{C}\mathit{a}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

 

1 mole of CaCl₂ produces 1 mole of Ca.

 

Now, moles of Ca

 

 $\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{1}}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\\ \mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\left(}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{\right)}\end{array}$

Hence, moles of Ca are 7.9454mol.

 

Step 6: Calculation of mass of Ca

Again you know,

$\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{=}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{\times }\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathit{\left(}\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{\right)}$ 

 

Molar mass of Ca is 40g/mol.

 

Now, mass of Ca

$\begin{array}{l}\mathbf{=}\mathbf{7}\mathbf{.9454}\mathbf{\times }\mathbf{40}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{=}\mathbf{317}\mathbf{.816}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\end{array}$ 

 

Hence, the other product is Ca.

 

317.816kg of Ca can be prepared from the electrolytic decomposition of 874g of CaCl₂.