Solution Stoichiometry

96) m_H2SO4 = 5.0 x 10³ kg H₂SO₄          D = 1.84 g/mL

      m%(H₂SO₄) = 95.0%                        m = ? Na₂CO₃

      H₂SO₄(aq) + Na₂CO₃(aq) Na₂SO₄(aq) + H₂O(l) + CO₂(g)

      2H⁺(aq) + CO₃^2-(aq) H₂O(l) + CO₂(g)

      m%(H₂SO₄) = m_H2SO4/m_sol x 100%

      m_H2SO4 = 0.950 x 5.0 x 10³ kg  H₂SO₄ x 10³ g/1 kg = 4.8 x 10⁶ g H₂SO₄  

      m_Na2CO3 = 4.8 x 10⁶ g H₂SO₄ x 1 mol H₂SO₄/98.08 g H₂SO₄ x 1 mol Na₂CO₃/1 mol H₂SO₄ x 105.99 g Na₂CO₃/ 1 mol Na₂CO₃ x 1 kg/10³g

      m_Na2CO3 = 5.2 x 10³ kg Na₂CO₃