ow many grams of water are formed when 45 moles of sulfuric acid react with 50 moles of sodium hydroxide? Which reactant is limiting? more in the description

Question

How many moles of the excess reactant will be left after the reaction?H2SO4 + 2 NaOH  Na2SO4 + 2 H2

Answer 1

(45 mol H₂SO₄) * (2 mol H₂O / 1 mol H₂SO₄) = 90 mol H₂O.

(50 mol NaOH) * (2 mol H₂O / 2 mol NaOH) = 50 mol H₂O. Sodium hydroxide is limiting, with 40 moles excess of sulfuric acid.

Mass of water: (50 mol H₂O) * (18.0g H₂O / 1 mol H₂O) = 900g H₂O.