how do you break out of a while loop after a certain amount of time ?

You could try this perhpas...

i = tick() --start time
while tick() < i+"DURATION IN SECONDS" do
wait()
--whatever

end

Tick() is the number of seconds passed since 01/01/1970 so by checking the current tick compared to the start time (i) + the duration you want it should create a timer without using wait(1) i = i+1 etc.

Not sure if this is the best way but I think it'll work!

So, you need to make a function that creates the tween, and plays it. Then, put the function in a script like this:

for I   = 1, 5, 1 do --[[ Note: The "5" is the amount of times you want it to loop. I    just put it there as an example]]--
-- Put function here
end

A simple but disadvantaged way you can do is this:

local Time = 0
while true do
    wait(0.1)   --can be changed.
    Time = Time + 1     --increases the time by 1.
    if Time == 15 then  -- 15 * 0.1 = 1.5 seconds. Again, can be changed.   
        break   --after 1.5 seconds it stops the while loop.
    end
end

I don't know what you're trying to do, but I'd recommend using a numeric for loop for this, it'll automatically stop when the given amount is reached.

for i = 1, 5, 1 do -- i is defined as the current number we are on, so it will chan each time, the first 1 is the start, the 5 is how many times we want it to do the loop and the other 1 is how many steps it takes.
     print(i);
end

Output: 1 2 3 4 5

I'll just point out the use of break with tick()

local curt = tick() -- current tick, when game starts
while true do wait()
    print("sup")
    if math.abs(curt - tick()) > 5 then
        break
    end
end

basically an original tick is made, a loop is formed, it prints sup a ton of times, if the last tick minus the current tick is greater than 5 seconds then break.

I guess you could just do:

for i = 10, 0, -1 --Numeric for loop that counts down from 10.
    wait()
end     

while i = 0 do
    wait()
    print(i.." seconds until loop breaks.")
end

Hope this helps!