Question #1a680

You know that an unknown reducing agent will reduce sodium nitrate to ammonia, so start by looking at the reduction half-reaction.

Keep in mind that you don't need to balance the half-reaction because you're only interested in how the oxidation state of nitrogen changes!

#stackrel(color(blue)(+5))("N") "O"_ 3""^(-) + 8"e"^(-) -> stackrel(color(blue)(-3))("N")"H"_ 3#

Now, notice that the oxidation state of nitrogen goes from #color(blue)(+5)# on the reactants' side to #color(blue)(-3)# on the products' side, which basically means that every atom of nitrogen gains #8# electrons.

This means that the #n# factor of the sodium nitrate, which acts as an oxidizing agent in this reaction, will be equal to #8#.

Now, in every redox reaction, the number of electrons lost in the oxidation half reaction must be equal to the number of electrons gained in the reduction half-reaction.

In your case, you know that #7.65 * 10^(-3)# moles of sodium nitrate will gain a total of

#7.65 * 10^(-3) color(red)(cancel(color(black)("moles NaNO"_3))) * "8 e"^(-)/(1color(red)(cancel(color(black)("mole NaNO"_3)))) = 6.12 * 10^(-2)# #"moles e"^(-)#

Since you know that the reaction consumed #3.06 * 10^(-2)# moles of this unknown reducing agent, you can say that you will have

#1 color(red)(cancel(color(black)("mole reducing agent"))) * (6.12 * 10^(-2)color(white)(.)"moles e"^(-))/(3.06 color(red)(cancel(color(black)("moles reducing agent")))) = "2 moles e"^(-)#

Since every #1# mole of this reducing agent loses #2# moles of electrons, you can say that its #n# factor will be equal to

#ncolor(white)(.)"factor reducing agent" = 2#

SIDE NOTE I would try to balance the reduction half-reaction like this--the reaction takes place in basic medium!

#9"H"_ 2"O" + stackrel(color(blue)(+5))("N") "O"_ 3""^(-) + 8"e"^(-) -> stackrel(color(blue)(-3))("N")"H"_ 3 + 3"H"_2"O" + 9"OH"^(-)#