Question #be953

1 Answer
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 #"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"#