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19 May, 15:47

What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride

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  1. 19 May, 19:01
    0
    Given:

    Volume of Na2CO3 = 250 ml = 0.250 L

    Molarity of Na2CO3 = 6.0 M

    Volume of CaF2 = 750 ml = 0.750 L

    Molarity of CaF2 = 1.0 M

    To determine:

    The mass of CaCO3 produced

    Explanation:

    Na2CO3 + CaF2 → CaCO3 + 2NaF

    Based on the reaction stoichiometry:

    1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

    Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

    Moles of CaF2 present = V * M = 0.750 * 1 = 0.750 moles

    CaF2 is the limiting reagent

    Thus, # moles of CaCO3 produced = 0.750 moles

    Molar mass of CaCO3 = 100 g/mol

    Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g

    Ans: Mass of CaCO3 produced = 75 g
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