What volume of 0.4567 m h2so4 is required to neutralize 30.00 ml of 0.3210 m naoh?

Answer is: volume of H₂SO₄ is 42.1 mL.
Chemical reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
c(H₂SO₄) = 0,4567 M = 0,4567 mol/L.
V(NaOH) = 30 mL ÷ 1000 mL/L = 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0,01926 mol.
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.
V(H₂SO₄) = 0,0421 L = 42,1 mL.

I’m not sure what your trying to say here.. can you clarify again?