How many mL of a 0.38 mol H2SO4 solution are needed to neutralize 280 mL of a 0.98 mol NaOH? Please explain how to work it out :(?

In the question, "0.38 mol" and "0.98 mol" should be "0.38 M" and "0.98 M" instead.

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
OR: Mole ratio H₂SO₄ : NaOH = 1 : 2

No. of moles of NaOH reacted = (0.98 mol/L) × (280/1000 L) = 0.2744 mol
No. of moles of H₂SO₄ needed = (0.2744 mol) × (1/2) = 0.1372 mol
Volume of H₂SO₄ needed = (0.1372 mol) / (0.38 mol/L) = 0.36 L = 360 mL

Calculate the moles of NaOH:
0.280 x 0.98 mole/L = 0.2744 mole of NaOH
Write the balanced equation:
H2SO$ + 2 NaOH --> Na2SO4 + 2 H2O
1 mole of H2SO4 will neutralize 2 mole of NaOH
You need 1/2 x 0.2744 mole of H2SO4 (0.1372 mole)
Calculate the liters of 0.38 M H2SO4 that contain 0.1372 mole.
L x 0.38 mole/L = 0.1372
? L = 0.361 L
Since you have only 2 significant figures for the molarity of NaOH and H2SO4, the answer should contain only 2 significant figures.
ANSWER: 0.36 L (360 mL)