Q86P - Chemistry: Molecular Nature Of Matter And Change

Question

When 56.6 g of calcium and 30.5 g of nitrogen gas undergo a reaction that has a 93.0% yield, what mass of calcium nitride forms?

Step-by-Step Solution

Answer

Actual yield or mass formed of  ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$during the reaction is 64.8 g.

Step 1: Writing balanced equation

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

 $\mathbf{3}\mathbf{Ca}\mathbf{+}{\mathbf{N}}_{\mathbf{2}}\mathbf{\to }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$

 

From the equation we can say that 3 moles of Careact with 1 mole of nitrogen gas to giveone mole of  ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$

Step 2: Calculating Moles of reactants

Moles can be calculated as 

 

 $\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ca}\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ca}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}}\mathbf{=}\frac{\mathbf{55}\mathbf{.}\mathbf{6}}{\mathbf{40}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{415}$

 

 $\mathbf{Mole}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }{\mathbf{N}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{ }\mathbf{Mass}}\mathbf{=}\frac{\mathbf{30}\mathbf{.}\mathbf{5}}{\mathbf{28}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{09}$

 

From the balanced equation we can say Ca is the limiting reagent hence controls the formation of products and,

 

 $\frac{\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ca}\mathbf{ }\mathbf{reacted}}{\mathbf{3}}\mathbf{=}\mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}\mathbf{formed}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{415}}{\mathbf{3}}\mathbf{0}\mathbf{.}\mathbf{47}\mathbf{mole}$

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 ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$ is 

 

 $\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}\mathbf{=}\mathbf{Mole}\mathbf{\times }\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{47}\mathbf{\times }\mathbf{148}\mathbf{.}\mathbf{25}\mathbf{=}\mathbf{69}\mathbf{.}\mathbf{67}\mathbf{g}$

Step 4: Actual yield calculation

Percent yield =93 %

Percent yield is calculated as:

 

 $\mathbf{Percent}\mathbf{ }\mathbf{yield}\mathbf{ }\mathbf{=}\frac{\mathbf{Actual}\mathbf{ }\mathbf{yield}}{\mathbf{theoritical}\mathbf{ }\mathbf{yield}}\mathbf{\times }\mathbf{100}\mathbf{=}\mathbf{93}$

 

 $\mathbf{Actual}\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{93}}{\mathbf{100}}\mathbf{\times }\mathbf{theoritical}\mathbf{ }\mathbf{yield}\mathbf{=}\frac{\mathbf{93}}{\mathbf{100}}\mathbf{\times }\mathbf{69}\mathbf{.}\mathbf{67}\mathbf{=}\mathbf{64}\mathbf{.}\mathbf{8}\mathbf{g}$

 

So, actual yield or mass formed of ${\mathbf{Ca}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$ during the reaction is 64.8 g.