Answer:
2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;
Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;
2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O
Explanation:
First, let's see the reactants for the first reaction and how they dissociate:
HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻
BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)
So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:
H⁺ + CO₃²⁻ → H₂CO₃
Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂
The reaction then is:
HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃
The number of elements must be the same on both sides, so the balanced equation is
2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃
The treatment with H₂SO₄ will produce:
H₂SO₄ → 2H⁺ + SO₄⁻²
Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻
The balanced reaction will be then:
Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃
In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂
HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻
Ca(OH)₂ → Ca²⁺ + 2OH⁻
The balance reaction will be:
2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O
