What mass of sodium carbonate would be required to remove essentially all of the calcium ion from 750 l of solution containing 43 mg ca2+ per liter?

Ca²⁺ reacts with CO₃²⁻ and forms CaCO₃. The reaction is
 CO₃²⁻(aq) + Ca²⁺(aq) → CaCO₃(s)
 
First, let's find out the moles of Ca²⁺ in 750 L.
     1 L has 43 mg of Ca²⁺
     Hence, mass of Ca²⁺ in 750 L = 43 mg/L x 750 L
                                                     = 32250 mg = 32.25 g
     Molar mass of Ca²⁺ = 40.078 g/mol
     Hence, moles of Ca²⁺ in 750 L = mass / molar mass
                                                        = 32.25 g / 40.078 g/mol
                                                        = 0.8047 mol

The stoichiometric ratio between Ca²⁺ and CO₃²⁻ is 1 :1
Hence, moles of CO₃²⁻ = moles of Ca²⁺
                                       = 0.8047 mol
 
CO₃²⁻ ions from dissociation of Na₂CO₃. The dissociation of Na₂CO₃ in water is
Na₂CO₃(s) → 2Na⁺(aq) + CO₃²⁻(aq)

The stoichiometric ratio between Na₂CO₃(s) and CO₃²⁻(aq) is 1 : 1
Hence, moles of Na₂CO₃(s) = moles of CO₃²⁻(aq)
                                               = 0.8047 mol
 
Molar mass of Na₂CO₃ = 105.9888 g/mol
Hence, mass of Na₂CO₃ = moles x molar mass
                                         = 0.8047 mol x 105.9888 g/mol
                                         = 85.29 g

Hence, mass of Na₂CO₃ needed to react with Ca²⁺ is 85.29 g.

The answer for this question is d. none of these

The answer is letter c