Can somebody please answer this

Question

A 12.5 mL sample of 0.25 M H2SO4 solution is completely neutralize by 0.150 M KOH solution. Calculate the volume of KOH solution used in this neutralization reaction

H₂SO_{4 (aq)} + KOH → K₂SO_{4 (aq)} +H₂O_(l)

Answer 1

H₂SO_{4 (aq)} + 2KOH ==> K₂SO_{4 (aq)} + 2H₂O_(l) ... balanced equation

-Calculate moles of H₂SO₄ used

12.5 ml x 1 L / 1000 ml x 0.25 mol /L = 0.003125 mols H₂SO₄ used

-Calculate moles of KOH needed to neutralize 0.003125 moles of H₂SO₄(use stoichiometry of balanced eq.)

0.003125 mols H₂SO₄ x 2 mol KOH / 1 mol H₂SO₄ = 0.00625 mols KOH

-Calculate the volume of KOH containing 0.00625 moles

0.00625 mols KOH x 1 L / 0.150 mols = 0.04167 L = 41.7 mls of KOH used in neutralization