i am completely lost, i confused myself.

i need to find the concentration of HC2H3O2

HC2H3O2 + NaOH ==> NaC2H3O2 + HOH

NaOH
C= 0.3 M
v= 0.25 L
n= 0.075 moles
M= 40.00g/mol
m= 3.00 g

NaC2H3O2
n= 0.075 moles
m= 0.4973 g(ammount of acetic acid in
solution)

HC2H302
v= 0.020 L
M= 60.06g/mol

thats all the information i have. i don't know how to calculate it. my friend said c1v1=c2v2 but i don't know what numbers id use

You are confused because you have too much information piled in together. You need to think this through. And you don't say what unit the concentration should be in.

M(NaOH) x L(NaOH) = mols NaOH.
mols NaOH = mols acetic acid since the equation is 1 mol NaOH to 1 mol acetic acid.

Now M(acid) x L(acid) = mols. You know mols, you know L acid (is that the 0.020 L), calculate M acid. That will be the molarity of the acetic acid in that 20 mL sample. By the way, I can't believe you used 0.25 L of the NaOH. That's 250 mL which is a huge volume to be titrated with. That means you would have had to refill a 50 mL buret five times. (Of course this could be a problem.)
I'm getting ready to leave for the night so if there is a quick question, please type it in before I leave.

alright thank you i think i should be able to figure it out. and i know we only did 3 trials though

To find the concentration of HC2H3O2, you can use the equation c1v1 = c2v2, where c1 is the initial concentration, v1 is the initial volume, c2 is the final concentration, and v2 is the final volume.

In this case, you are given the concentration and volume of NaOH (c = 0.3 M and v = 0.25 L) and the volume of HC2H3O2 (v = 0.020 L). The goal is to find the concentration of HC2H3O2 (c2).

Since the reaction between NaOH and HC2H3O2 is a 1:1 stoichiometry (1 mole of NaOH reacts with 1 mole of HC2H3O2), the number of moles of NaOH is the same as the number of moles of HC2H3O2.

First, let's find the number of moles of NaOH using the given mass and molar mass:
n(NaOH) = m / M = 3.00 g / 40.00 g/mol = 0.075 moles

Since the stoichiometry is 1:1, the number of moles of HC2H3O2 is also 0.075 moles.

Now, using the equation c1v1 = c2v2, plug in the known values:
(0.3 M)(0.25 L) = c2(0.020 L)

Solving for c2:
c2 = (0.3 M)(0.25 L) / (0.020 L) = 3.75 M

Therefore, the concentration of HC2H3O2 is 3.75 M.