Calculate the mass of water produced when 0.333L of 0.500M NaOH is added to 3.0 grams of acetic acid.

The equation is: NaOH(aq) + HC2H302(aq) -> NaC2H3O2 (aq) + H20(l)

Since we have 0.333 L of 0.500 M NaOH, we have:
(0.500 mol NaOH)/(1 L NaOH) * (0.333 L NaOH) = 1.665 mol NaOH.

Then, since we have 3.0 g of HC2H3O2, which has a molar mass of 1.01 + 2(12.01) + 3(1.01) + 2(16) = 60.06 g/mol, we have:
(3.0 g HC2H3O2) * (1 mol HC2H3O2)/(60.06 g HC2H3O2) = 5.00 x 10^-2 mol HC2H3O2.

It shouldn't be to any surprise that HC2H3O2 is the limiting reactant. Since water is produced in a 1:1 ratio, we also have 5.00 x 10^-2 mol of water. Since the molar mass of water (H2O) is 2(1.01) + 16 = 18.02 g/mol, we have:
(5.00 x 10^-2 mol H2O) * (18.02 g H2O)/(1 mol H2O) = 0.90 g H2O.

I hope this helps!