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17 January, 08:59

A mixture of calcium carbonate, CaCO3, and barium carbonate, BaCO3, weighing 5.40 g reacts fully with hydrochloric acid, HCl (aq), to generate 1.39 L CO2 (g), mea sured at 50°C and 0.904 atm pressure. Calculate the percentages by mass of CaCO3 and BaCO3 in the original mixture.

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  1. 17 January, 11:20
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    CaCO₃ = 85.18%

    BaCO₃ = 14.82%

    Explanation:

    The acid will react with the salts, the partial reactions are:

    CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

    BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

    So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

    Using the ideal gas law to calculate the number of moles of CO₂:

    PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

    0.904*1.39 = n*0.082*323

    26.486n = 1.25656

    n = 0.05 mol

    So, the number of moles of the mixture is 0.05 mol.

    The molar masses of the components are:

    CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

    BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

    Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

    100x + 197.3y = 5.4

    x + y = 0.05 mol

    y = 0.05 - x

    100x + 197.3 * (0.05 - x) = 5.4

    100x - 197.3x = 5.4 - 9.865

    97.3x = 4.465

    x = 0.046 mol of CaCO₃

    y = 0.004 mol of BaCO₃

    So, the masses are:

    CaCO₃ = 100 * 0.046 = 4.60 g

    BaCO₃ = 137.3*0.004 = 0.80 g

    The percentages in the mixture are:

    CaCO₃ = (4.60/5.40) * 100% = 85.18%

    BaCO₃ = (0.80/5.40) * 100% = 14.82%
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