If 31 mL of 5.6 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?

Answer 1

H₂SO₄ + 2NaHCO₃ ==> Na₂SO₄ + 2CO₂ + 2H₂O ... balanced equation

mols of H₂SO₄ spilled = 31 ml x 1 L / 1000 ml x 5.6 mol/L = 0.1736 mols

mols NaHCO₃ needed = 0.1736 mols H₂SO₄ x 2 mol NaHCO₃/1 mol H₂SO₄ = 0.3472 mols NaHCO₃

mass NaHCO₃ needed = 0.3472 mols x 84.0 g/mol = 29 g NaHCO₃ (2 sig. figs.)