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22 July, 10:56

If 12.8 g of CaCO3 decomposes at 38 degrees C and 0.96 atm, how many dm3 of CO2 are formed in addition to CaO? CaCO3 → CaO + CO2

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  1. 22 July, 14:52
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    3.4dm3

    Explanation:

    We'll begin by calculating the number of mole of CaCO3 present in 12.8g of CaCO3.

    This can be achieved as shown below:

    Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

    Mass of CaCO3 obtained from the question = 12.8g

    Number of mole of CaCO3 = ?

    Number of mole = Mass / Molar Mass

    Number of mole of CaCO3 = 12.8/100

    Number of mole of CaCO3 = 0.128 mole

    The equation for the reaction is given below:

    CaCO3 → CaO + CO2

    From the equation above,

    1 mole of CaCO3 produced 1 mole of CO2.

    Therefore, 0.128 mole of CaCO3 will also produce 0.128 mole of CO2.

    Now, we can obtain the volume of CO2 produced as follow:

    Data obtained from the question include:

    T (temperature) = 38°C = 38 + 273 = 311K

    P (pressure) = 0.96 atm

    n (number of mole of CO2) = 0.128 mole

    R (gas constant) = 0.082atm. dm3/Kmol

    V (volume of CO2) = ?

    Using the ideal gas equation PV = nRT, the volume of CO2 produced can be obtained as shown:

    PV = nRT

    0.96 x V = 0.128 x 0.082 x 311

    Divide both side by 0.96

    V = (0.128 x 0.082 x 311) / 0.96

    V = 3.4dm3

    Therefore, 3.4dm3 of CO2 are produced.
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