Balancing redox equation with oxygen on both sides?
It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are:
$$ \begin{align}
\ce{As2O3 (s) -&gt; H3AsO4 (aq)} \\
\ce{NO3- (aq) -&gt; N2O3 (aq)}\\
\end{align}
$$

First balance $\ce{As}$ and $\ce{N}$ in both equations. Then, balance $\ce{O}$ in both equations with water and $\ce{H}$ with $\ce{H+}$:

$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -&gt; 2 H3AsO4 (aq) + 4 H+ (aq)}\\
\ce{2 NO3- (aq) + 6 H+ (aq) -&gt; N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Finally, balance the charges by electrons:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -&gt; 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\
\ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -&gt; N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $\ce{e-}$s):

$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -&gt; 2 H3AsO4 (aq) + N2O3 (aq)}$$
Or simply,
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq)  -&gt; 2 H3AsO4 (aq) + N2O3 (aq)}$$