# What volume of nitric acid of 5.0*mol*L^-1 concentration is required to oxidize a 16*g mass of Fe^(2+) ions?

Apr 24, 2017

Approx. $76.5 \cdot m L$.........

#### Explanation:

$F {e}^{2 +}$ is oxidized to $\text{ferric ion}$:

${\text{Fe"^(2+) rarr "Fe}}^{3 +} + {e}^{-}$ $\left(i\right)$

And nitric acid is reduced to nitrous oxide:

$\text{H"stackrel(V+)"NO"_3 + "3H"^(+) + 3e^(-) rarrstackrel(II+)"NO" +"2H"_2"O}$ $\left(i i\right)$

And so $3 \times \left(i\right) + \left(i i\right)$ gives............

$3 F {e}^{2 +} + 4 H N {O}_{3} \rightarrow 3 F {e}^{3 +} + N O \left(g\right) \uparrow + 3 N {O}_{3}^{-} + 2 {H}_{2} O \left(l\right)$

I have treated ${H}^{+} \equiv H N {O}_{3}$ here...........

Now you have claimed you got $16 \cdot g$ of ferrous iron (which is less than the equivalent mass of ferrous salt, like $F e S {O}_{4}$.

Going with $16 \cdot g$, this represents a molar quantity of........

$\frac{16 \cdot g}{55.8 \cdot g \cdot m o {l}^{-} 1} = 0.287 \cdot m o l$, and thus we need $\frac{4}{3} \times 0.287 \cdot m o l = 0.382 \cdot m o l$ with respect to nitric acid.

And finally, $\text{moles"="concentration"xx"volume}$, or $\text{volume"="moles"/"concentration}$,

$= \frac{0.382 \cdot m o l}{5.0 \cdot m o l \cdot {L}^{-} 1} = 76.5 \cdot m L$.......

Remember I have calculated with respect to a mass of ferrous iron, not to a mass of ferrous salt.