Powering a self-drive piezo buzzer with capacitors?
Guessing without a datasheet is not the best. You can expect that to use around 25mA give or take 10 or 20mA. Let's say 30mA and 30 seconds and you will power it at 6V and allow it to fall to 4V (average voltage 5V and assumed constant 30mA, which won't be true, but is close enough for a WAG without a datasheet). 

So \$C = \frac {2(0.03 \cdot 5 \cdot 30)}{V_f^2-V_I^2}\$ = 0.45F

Since the energy consumed is the integral of i(t) * v(t) over time, and the energy in a capacitor is \$ \frac{C V^2}{2}\$