What volume of 0.500 m h2so4 is needed to react completely with 20.0 ml of 0.400 m lioh?

The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol 
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol 
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol  in 1 L solution 
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L 
therefore volume of acid required = 8 mL