Question

Question #be953

Answer 1

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 `"N"_2`.
3. Use the molar ratio of `"product":"N"_2` from the balanced equation to calculate the moles of `"product"`.
4. Use the molar mass of the `"product"` to calculate its mass.

Let's get started.

Step 1. Write the balanced chemical equation.

`M_text(r):color(white)(mmmmmmm)82.98`
`color(white)(mm)"6Na + N"_2 → "2Na"_3"N"`

Step 2. Calculate the moles of `"N"_2`.

The Ideal Gas Law is

`color(blue)(bar(ul(|color(white)(a/a)pV= nRTcolor(white)(a/a)|)))" "`

We can rearrange the Ideal Gas Law to get

`n = (pV)/(RT)`

In this problem,

`P = "1 bar"`
`V = "45.8 L"`
`R = "0.083 14 bar·L·K"^"-1""mol"^"-1"`
`T = 0^@"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 `"Na"_3"N"`.

`"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 `"Na"_3"N"`.

`"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"`