Balancing the oxidative decomposition of FeS2
I found a solution which works, 
$$\ce{4FeS_2 + 15O_2 +14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$



This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand. 

So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction : 

$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$

Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :

$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\  \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

It can be easily transform into a system like this :

$$\begin{cases}
2c+3d+2e=0 \\
2b+c+3d+4e=0 \\
2a+e=0 \\
a+d=0
\end{cases}$$

So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.

So here we will determine four unknowns in terms of an other, I chose $e$. And I got : 
$$$$\begin{cases}
a=-e/2 \\
b=-15e/8 \\
c=-7e/4 \\
d=e/2 \\
e=e
\end{cases}$$$$

I multiply then by $8$ to have only integers which give me with $e=-1$ :

$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$

Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$