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17 May, 01:41

You placed 6.35 g of a mixture containing unknown amounts of BaO (s) and MgO (s) in a 3.50-L flask containing CO₂ (g) at 30.0°C and 750. torr. The solid BaO and MgO in the flask completely reacted to form BaCO₃ (s) and MgCO₃ (s), respectively. After the reactions to form BaCO₃ (s) and MgCO₃ (s) were completed, the pressure of CO₂ (g) remaining was 215 torr, still at 30.0°C. Calculate the mass of BaO (s) in the initial mixture. (Assume ideal gas behavior).

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  1. 17 May, 02:07
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    Mass of BaO in initial mixture = 3.50g

    Explanation:

    Let mass of BaO in mixture be x g

    mass of MgO in mixture be (6.35 - x) g

    Initially CO_2

    Volume = 3.50 L

    Temp = 303 K

    Pressure = 750 torr = 750 / 760 atm

    Applying ideal gas equation

    PV = nRT

    n = PV / RT

    (n) _CO_2 = ((750/760) * 3.50) / 0.0821 * 303

    (n) _CO_2 = 0.139 mole

    Finally; mole of CO_2

    n = PV / RT

    ((245/760) * 3.5) / 303 * 0.0821

    (n) _CO_2 = 0.045 mole

    Mole of CO_2 reacted = 0.139 - 0.045

    =0.044 mole

    BaO + CO_2 BaCO_3

    Mgo + CO_2 MgCO_3

    moles of CO_2 reacted = (moles of BaO + moles of MgO)

    moles of BaO in mixture = x / 153 mole

    moles of MgO in mixture = 6.35 - x mole / 40

    Equating,

    x / 153 + 6.35/40 = 0.094

    = x/153 + 6.35 / 40 - x/40 = 0.094

    = x (1/40 - 1153) = (6.35/40 - 0.094)

    = x * 10.018464

    = 0.06475

    mass of BaO in mixture = 3.50g
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