Chapter 16, Problem 15STP

Interpretation Introduction

Interpretation:

The amount of nitrogen gas that can be produced from the $32.5$ grams of the sodium nitride needs to be determined.

Concept introduction:

The reaction of ${\text{Na}}_{\text{3}}\text{N}$ to produce Na metal and ${\text{N}}_{\text{2}}\left(\text{g}\right)$ is a decomposition reaction. Moles number of any substance is calculated by diving given mass by molar mass of the compound expressed in grams.

Answer to Problem 15STP

Amount of nitrogen gas produced from the $32.5$ grams of the sodium nitride is $5.49\text{g}$.

Explanation of Solution

Sodium nitride broken down to produce Na metal and ${\text{N}}_{\text{2}}\left(\text{g}\right)$. The balanced chemical equation for this reaction is-

${\text{2Na}}_{\text{3}}\text{N}\to 6{\text{Na+N}}_{\text{2}}\left(\text{g}\right)$

Calculate moles number of sodium nitride in $32.5$ grams as follows:

Molar mass of ${\text{Na}}_{\text{3}}\text{N}$ is $83.0\text{g/mol}$, mass of the ${\text{Na}}_{\text{3}}\text{N}$ is $32.5$ g.

$\begin{array}{c}\text{moles}\text{number}\text{of}{\text{Na}}_{\text{3}}\text{N=}\frac{\text{Given}\text{mass}\text{of}{\text{Na}}_{\text{3}}\text{N}}{\text{molar}\text{mass}\text{of}{\text{Na}}_{\text{3}}\text{N}}\\ =\frac{32.5\text{g}}{83.0\text{g/mol}}\\ \text{=0}\text{.392}\text{mol}\end{array}$

Therefore, number of moles present in $32.5$ g of ${\text{Na}}_{\text{3}}\text{N}$ is $\text{0}\text{.392}\text{mol}$.

Now, determine how much ${\text{N}}_{\text{2}}$ is $\text{0}\text{.392}\text{mol}$. From the balanced equation the mole ratio is $\frac{\text{1}\text{mol}{\text{N}}_{\text{2}}}{\text{2}\text{mol}{\text{Na}}_{\text{2}}{\text{N}}_{\text{2}}}$. Therefore, multiply the moles number of ${\text{Na}}_{\text{3}}\text{N}$ by the mole ratio.

$\begin{array}{c}\left(0.392\text{mol}{\text{Na}}_{\text{2}}\text{N}\right)\left(\frac{\text{1}\text{mol}{\text{N}}_{\text{2}}}{\text{2}\text{mol}{\text{Na}}_{\text{2}}{\text{N}}_{\text{2}}}\right)=\frac{0.392\times \text{1}\text{mol}{\text{N}}_{\text{2}}}{2}\\ =0.196\text{mol}{\text{N}}_{\text{2}}\end{array}$

So, the moles number of ${\text{N}}_{\text{2}}$ gas produced is $0.196\text{mol}$.

Convert moles number of ${\text{N}}_{\text{2}}$ gas into grams as follows:

Molar mass of N is 28 g/mol.

Therefore,

$\begin{array}{c}\text{mass}\text{of}{\text{N}}_{\text{2}}=0.196\text{mol}\frac{28\text{g}}{\text{mol}}\\ =5.49\text{g}\end{array}$

Amount of nitrogen gas that can be produced from the $32.5$ grams of the sodium nitride is $5.49\text{g}$.

Conclusion

Amount of nitrogen gas produced from the $32.5$ grams of the sodium nitride is $5.49\text{g}$.