Final answer:
To react completely with 72.5 g of sodium hydroxide, we would need 88.9 grams of sulfuric acid.
Step-by-step explanation:
In a neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH), the balanced equation is:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
To determine the number of grams of sulfuric acid needed to react completely with 72.5 g of sodium hydroxide, we need to first calculate the number of moles of sodium hydroxide using its molar mass (40.00 g/mol):
Moles of NaOH = Mass of NaOH / Molar mass of NaOH = 72.5 g / 40.00 g/mol = 1.81 mol NaOH
Since the mole ratio between H₂SO₄ and NaOH is 1:2, we can use that to calculate the number of moles of sulfuric acid needed:
Moles of H₂SO₄ = 1/2 * Moles of NaOH = 1/2 * 1.81 mol = 0.905 mol H₂SO₄
Finally, we can convert the moles of sulfuric acid to grams using its molar mass (98.09 g/mol):
Mass of H₂SO₄ = Moles of H₂SO₄ * Molar mass of H₂SO₄ = 0.905 mol * 98.09 g/mol = 88.9 g H₂SO₄