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