Percent Yield Problem!!!?

What is the percent yield if 155 g of calcium carbonate is treated with 250g of anhydrous hydrogen chloride and only 142 g of CaCl2 is obtained?

1 Answer

  • 8 years ago
    Favorite Answer

    First you need the molar masses of HCl, CaCl2, CaCO3

    HCl: 36.46 g/mol

    CaCl2: 110.98 g/mol

    CaCO3: 100.09 g/mol

    The equation for the reaction is:

    CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)

    First convert 155g of calcium carbonate into moles.

    155g CaCO3 / 100.09 g/mol = 1.5486 mol CaCO3

    Then do the same for your HCl

    250g HCl / 36.46 g/mol = 6.8568 mol HCl

    Then apply them using the equation to find the limiting reactant.

    1.5486 mol CaCO3 x 1 mol CaCl2 per 1 mol CaCO3 = 1.5486 mol CaCl2

    6.8568 mol HCl x 1 mol CaCl2 per 2 mol HCl = 3.4284 mol CaCl2

    Since CaCO3 produces the lower value, it is the limiting reactant, and the HCl is in excess. Therefore, we know the theoretical yield of CaCl2 is 1.5486 moles. Now we convert this into grams.

    1.5486 mol CaCl2 x 110.98 g/mol = 171.86g CaCl2

    The percentage yield is simply (actual yield / theoretical yield) x 100

    (142 / 171.86) x 100

    = 82.6%

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