Respuesta :

Ishankahps
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.

Answer:

Mass of Na2CO3 = 85.3 g

Explanation:

Volume of solution, V = 750 L

Mass of Ca2+/L = 43 mg

Mass of Ca2+  = 40.08 g

Mass of Na2CO3 = 105.99 g/mol

The net ionic equation between Ca2+ and CO3^2- is:

[tex]Ca^{2+} + CO3^{2-}  ------ CaCO_{3}[/tex]

Based on the stoichiometry:

1 mole of Ca2+ ions combines with 1 mole of CO₃²⁻ ions

Step 1: Calculate moles of Ca2+ in 750 L solution

Mass of Ca2+ in 750 L = [tex]\frac{0.043 g * 750 L}{1 L} = 32.25 g[/tex]

Moles of Ca2+ = [tex]\frac{32.25 g}{40.08 g/mol} =0.805 moles[/tex]

Step 2: Calculate the mass of Na2CO3 required

Moles of Ca2+ = Moles of Co3^2- = 0.805 moles

Mass of Na2CO3 = [tex]0.805 moles * 105.99 g/mol = 85.32 g[/tex]

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