Question #d4721

1 Answer
Aug 16, 2016


#2.4 * N_A#


The first thing to do here is to identify how many valence electrons you have in one nitride anion, #"N"^(3-)#.

As you know, a neutral nitrogen atom has #5# valence electrons. When the nitride anion is formed, #3# electrons are added to nitrogen's valence shell.

This means that a single nitride anion will have a total of

#underbrace("5 e"^(-))_ (color(blue)("neutral atom")) + underbrace("3 e"^(-))_ (color(purple)("added electrons")) = underbrace("8 e"^(-))_ (color(darkgreen)("nitride anion"))#

The next thing to do here is to figure out how many moles of nitride anions you have in your sample. To do that, use the molar mass of nitrogen, #"N"#.

This works because the molar mass of an ion is practically identical to the molar mass of the neutral atom.

#4.2 color(red)(cancel(color(black)("g"))) * "1 mole N"^(3-)/(14.007 color(red)(cancel(color(black)("g")))) = "0.29985 moles N"^(3-)#

Now all you have to do is figure out how many nitride anions you have in that many moles. As you know, Avogadro's number gives you the exact number of particles present in one mole.

In this case, you know that one mole of nitride anions contains #N_A# nitride anions. This means that #0.29985# moles will contain

#0.29985 color(red)(cancel(color(black)("moles N"^(3-)))) * (N_Acolor(white)(a)"N"^(3-)"anions")/(1color(red)(cancel(color(black)("mole N"^(3-))))) = 0.29985 * N_Acolor(white)(a)"N"^(3-)"anions"#

Finally, since you know that one nitride anion contains #8# valence electrons, you can say that you have

#0.29985 * N_A color(red)(cancel(color(black)("N"^(3-)"anions"))) * "8 valence e"^(-)/(1color(red)(cancel(color(black)("N"^(3-)"anion")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2.4 * N_A " valence e"^(-))color(white)(a/a)|)))#

The answer is rounded to two sig figs.