There are two different vats. The first vat has 20 gallons is being filled at a rate of 1.51.gallons per minute. The second vat has 120 gallons and is leaking fluid at a rate of 2.5 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Let \( t \) be the time in minutes after which both vats will contain the same amount of fluid.

For the first vat:
- It starts with 20 gallons and is filling at 1.51 gallons per minute.
- Therefore, the amount of fluid in the first vat after \( t \) minutes, \( V_1(t) \), is:
\[
V_1(t) = 20 + 1.51t
\]

For the second vat:
- It starts with 120 gallons and is leaking at a rate of 2.5 gallons per minute.
- Therefore, the amount of fluid in the second vat after \( t \) minutes, \( V_2(t) \), is:
\[
V_2(t) = 120 - 2.5t
\]

We want to find \( t \) such that both vats have the same amount of fluid:
\[
V_1(t) = V_2(t)
\]

Setting the equations equal to each other:
\[
20 + 1.51t = 120 - 2.5t
\]

Now, combine terms:
\[
1.51t + 2.5t = 120 - 20
\]
\[
4.01t = 100
\]

Now, solve for \( t \):
\[
t = \frac{100}{4.01} \approx 24.94 \text{ minutes}
\]

To convert minutes to seconds:
\[
t \approx 24.94 \times 60 \approx 1496.4 \text{ seconds}
\]

Thus, after approximately \( \boxed{1496.4} \) seconds, both vats will have the same amount of liquid.