One of the half-reactions for the electrolysis of water is
$$
2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-}
$$
If $0.076 \mathrm{~L}$ of $\mathrm{O}_{2}$ is collected at $25^{\circ} \mathrm{C}$ and $755 \mathrm{mmHg}$, how many moles of electrons had to pass through the solution?