The reaction between sulfuric acid HSO and sodium hydroxide NaOH is an acid-base reaction, which produces water HO and sodium sulfate NaSO as products. The balanced chemical equation for this reaction is:HSO + 2NaOH NaSO + 2HOTo calculate the amount of sulfuric acid needed to neutralize 20 grams of sodium hydroxide, we will use stoichiometry. First, we need to find the molar masses of NaOH and HSO:NaOH: 22.99 Na + 15.999 O + 1.007 H = 39.996 g/molHSO: 2 1.007 H + 32.06 S + 4 15.999 O = 98.078 g/molNow, we can find the moles of NaOH in 20 grams:moles of NaOH = 20 g / 39.996 g/mol = 0.5001 molFrom the balanced equation, we can see that 1 mole of HSO reacts with 2 moles of NaOH. Therefore, we need half the moles of HSO to neutralize the given amount of NaOH:moles of HSO = 0.5001 mol / 2 = 0.25005 molFinally, we can find the mass of HSO needed:mass of HSO = 0.25005 mol 98.078 g/mol = 24.523 gSo, approximately 24.523 grams of sulfuric acid are needed to neutralize 20 grams of sodium hydroxide.