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?

Short answer

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

Step-by-step solution

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

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,

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

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

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

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.