Question 1a680

Jul 20, 2017

$n$ $\text{factor} = 2$

Explanation:

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 $\textcolor{b l u e}{+ 5}$ on the reactants' side to $\textcolor{b l u e}{- 3}$ 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 $7.65 \cdot {10}^{- 3}$ 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) ${\text{moles e}}^{-}$

Since you know that the reaction consumed $3.06 \cdot {10}^{- 2}$ 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

$n \textcolor{w h i t e}{.} \text{factor reducing agent} = 2$

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

$9 {\text{H"_ 2"O" + stackrel(color(blue)(+5))("N") "O"_ 3""^(-) + 8"e"^(-) -> stackrel(color(blue)(-3))("N")"H"_ 3 + 3"H"_2"O" + 9"OH}}^{-}$