3.63P - Chemistry: Molecular Nature Of Matter And Change

Question

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV):

$\mathbf{4}\mathbf{HCl}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{MnO}}_{\mathbf{2}}\left(s\right)\mathbf{\to }{\mathbf{MnCl}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\left(g\right)$ 

When 1.82 mol of $\mathbf{HCl}$ reacts with excess ${\mathbf{MnO}}_{\mathbf{2}}$ 

(a) How many moles of ${\mathbf{Cl}}_{\mathbf{2}}$ react?

(b) How many grams of ${\mathbf{Cl}}_{\mathbf{2}}$ form?

Step-by-Step Solution

Answer

1. The number of moles of  ${\mathbf{Cl}}_{\mathbf{2}}$ formed is 0.91 mol.
2. The mass of  ${\mathbf{Cl}}_{\mathbf{2}}$ formed is 64.5 g.

Step 1: Determine the Limiting reagent of the reaction

In the reaction, ${\mathrm{MnO}}_2$ is present in excess, so when 1.82 mol of $\mathrm{HCl}$ reacts with an excess of ${\mathrm{MnO}}_2,\mathrm{HCl}$ will be completely consumed in the reaction. Therefore, $\mathrm{HCl}$ is a limiting reagent.

This states that the reactant which is completely utilized in the reaction or completely used up in the reaction acts as a Limiting reagent.

Step 2: Relation between the number of moles of Limiting reagent and Cl 2 formed

In the given reaction, 4 mol of $\mathrm{HCl}$ reacts with ${\mathrm{MnO}}_2$ and forms 2mol of ${\mathrm{MnCl}}_2$ 

Thus,

$\begin{array}{rcl}4\mathrm{molofHCl}& =& 2{\mathrm{molofCl}}_2\\ 1\mathrm{molofHCl}& =& \frac{2{\mathrm{molofCl}}_2}{4}\\ & =& 0.5{\mathrm{molofCl}}_2\\ & & \end{array}$ 

Therefore, 1mol of $\mathrm{HCl}$ reacts and formed 0.5 mol of ${\mathrm{Cl}}_2$ 

 

Step 3: Calculate the number of moles of formed

The number of moles of ${\mathbf{Cl}}_{\mathbf{2}}$  is:

 $\begin{array}{rcl}1\mathrm{molofHCl}& =& 0.5{\mathrm{molofCl}}_2\\ 1.82\mathrm{molofHCl}& =& 1.82\times 0.5{\mathrm{molofCl}}_2\\ & =& 0.91{\mathrm{molofCl}}_2\\ & & \end{array}$

Hence, the number of moles of  ${\mathbf{Cl}}_{\mathbf{2}}$ formed is 0.91 mol.

Step 4: Relation between mass and number of moles

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

 $\mathrm{Number} \mathrm{of} \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}$

Step 5: Calculate the mass of the formed

The molar mass of ${\mathbf{Cl}}_{\mathbf{2}}$ is 70.906 g/mol.

Thus, the mass of ${\mathbf{Cl}}_{\mathbf{2}}$ is:

$\begin{array}{rcl}\mathrm{mass} \mathrm{of} {\mathrm{Cl}}_2& =& \mathrm{Number} \mathrm{of} \mathrm{moles}\times \mathrm{Molar} \mathrm{mass}\\ & =& 70.906 \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.\times 0.91\mathrm{mol}\\ & =& 64.52446\mathrm{g}\\ & =& 64.5\mathrm{g}\end{array}$ 

Therefore, the mass of ${\mathbf{Cl}}_{\mathbf{2}}$ is 64.5 g.