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23 September, 09:18

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?

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  1. 23 September, 10:31
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    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.
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