Trying to find the vertical acceleration by subtracting weight from lift
Your math looks right to me. If an airplane has the statistics that you listed there, then it will accelerate upwards at $16.48\ \mathrm{m}/\mathrm{s}^2$, as you calculated. The occupants will experience about 3 g's of proper acceleration (1 g from gravity, and about 2 g's from the acceleration).

But here's what would actually happen, in context. 

To start, assume that the plane is in straight and level flight at 33 m/s. (Here's why I'm making this assumption: You're subtracting weight from lift, which only makes sense if the lift vector is pointing straight up. And if the lift vector is pointing straight up, then the plane is probably in straight and level flight.)

While the plane is in straight and level flight, the lift must equal the weight, so the coefficient of lift must be much less than the 1.5 you have here. 

Then, the pilot suddenly pulls back on the yoke, causing the plane to pitch up. This causes the angle of attack to quickly increase, which causes the coefficient of lift to increase to 1.5. The plane starts accelerating upwards at $16.48\ \mathrm{m}/\mathrm{s}^2$.

At this point, let's assume that the pilot maintains a constant pitch after this initial pull-up.

As the plane's vertical speed increases, its climb angle also increases, which causes the angle of attack to decrease. This, in turn, causes the coefficient of lift to decrease again, until the lift and weight approximately balance out again. At this point, the plane is in a steady climb. The vertical acceleration is now $0\ \mathrm{m}/\mathrm{s}^2$, and the occupants are feeling just 1 g again.

(I'm making a couple of small simplifying assumptions in the above: I'm assuming that there is no updraft or downdraft, and that plane's engine is not producing a significant amount of upward force.)