Q85P - Chemistry: Molecular Nature Of Matter And Change

Question

When 20.5 g of methane and 45.0 g of chlorine gas undergo a reaction that has a 75.0% yield, what mass of chloromethane  $\left({\mathrm{CH}}_3\mathrm{Cl}\right)$forms? Hydrogen chloride also forms.

Step-by-Step Solution

Answer

Actual yield or mass formed of   ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$during the reaction is 24 g.

Step 1: Writing balanced equation

First of all, let us check the balanced equation to find the actual yield:

 ${\mathbf{CH}}_{\mathbf{4}}\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\to }{\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}\mathbf{+}\mathbf{HCl}$

From the equation we can say that 1 mol of ${\mathbf{CH}}_{\mathbf{4}}$ reacts with 1 mole of chlorine gas to giveone mole of  ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$

Step 2: Calculating Moles of reactants

Moles can be calculated as 

 $\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CH}}_{\mathbf{4}}\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }{\mathbf{CH}}_{\mathbf{4}}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}}\mathbf{=}\frac{\mathbf{20}\mathbf{.}\mathbf{5}}{\mathbf{16}}\mathbf{1}\mathbf{.}\mathbf{28125}$

 

$\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{Cl}}_2\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{Cl}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}}\mathbf{=}\frac{\mathbf{45}}{\mathbf{71}}\mathbf{0}\mathbf{.}\mathbf{634}$

From the balanced equation we can say  ${\mathbf{Cl}}_{\mathbf{2}}$is the limiting reagent hence controls the formation of products and,

 $\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{Cl}}_2\mathbf{ }\mathbf{reacted}\mathbf{=}\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CH}}_{\mathbf{3}}\mathbf{ }\mathbf{Cl}\mathbf{ }\mathbf{formed}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{634}$

Step 3: Theoretical yield

The maximum possible mass of a product that can be formed in a chemical reaction, is known as its theoretical yield. Hence, Theoretical yield of ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$ is:

 $\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}\mathbf{=}\mathbf{Mole}\mathbf{\times }\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{634}\mathbf{\times }\mathbf{50}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{32}\mathbf{g}$

Step 4: Actual yield calculation

Percent yield =75 %

Percent yield is calculated as:

 $$\begin{gathered} \mathbf{Percent}\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{Actual}\mathbf{ }\mathbf{yield}}{\mathbf{theoritical}\mathbf{ }\mathbf{ }\mathbf{yield}}\mathbf{\times }\mathbf{100}\mathbf{=}\mathbf{75} \\ \mathbf{Actual}\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{75}}{\mathbf{100}}\mathbf{\times }\mathbf{theoritical}\mathbf{ }\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{75}}{\mathbf{100}}\mathbf{\times }\mathbf{32}\mathbf{=}\mathbf{24}\mathbf{g} \end{gathered}$$

 

So, actual yield or mass formed of ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$ during the reaction is 24 g.