# Question be953

May 1, 2017

#### Answer:

The mass of the product is 335 g.

#### Explanation:

We aren’t given the temperature or pressure of the gas, so I will assume STP (1 bar and 0 °C).

There are four steps involved in this problem:

1. Write the balanced equation for the reaction.
2. Use the Ideal Gas Law to calculate the moles of ${\text{N}}_{2}$.
3. Use the molar ratio of ${\text{product":"N}}_{2}$ from the balanced equation to calculate the moles of $\text{product}$.
4. Use the molar mass of the $\text{product}$ to calculate its mass.

Let's get started.

Step 1. Write the balanced chemical equation.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m m m m} 82.98$
$\textcolor{w h i t e}{m m} \text{6Na + N"_2 → "2Na"_3"N}$

Step 2. Calculate the moles of ${\text{N}}_{2}$.

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange the Ideal Gas Law to get

$n = \frac{p V}{R T}$

In this problem,

$P = \text{1 bar}$
$V = \text{45.8 L}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = {0}^{\circ} \text{C = 273.15 K}$

n = (PV)/(RT) = (1 color(red)(cancel(color(black)("bar"))) × 45.8 color(red)(cancel(color(black)("L"))))/("0.083 14"color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K")))) = "2.017 mol"#

2. Calculate the moles of $\text{Na"_3"N}$.

$\text{Moles of Na"_3"N" = "2.017" color(red)(cancel(color(black)("mol N"_2))) × ("2 mol Na"_3"N")/(1 color(red)(cancel(color(black)("mol N"_2)))) = "4.034 mol Na"_3"N}$

3. Calculate the mass of $\text{Na"_3"N}$.

$\text{Mass of Na"_3"N" = "4.034" color(red)(cancel(color(black)("mol Na"_3"N"))) × ("82.98 g Na"_3"N")/(1 color(red)(cancel(color(black)("mol Na"_3"N")))) = "335 g Na"_3"N}$