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26 January, 01:06

If 0.158 g of a white, unknown solid carbonate of a group 2A metal (M) is heated and the resulting CO2 is transferred to a 285 ml sealed flask and allowed to cool to 25 degrees Celsius, the pressure in the flask is 69.8 mmHg. What is the identity of the carbonate?

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  1. 26 January, 02:20
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    Using the ideal gas law equation we can find the number of moles of CO₂ formed

    PV = nRT

    where

    P - pressure - 69.8 mmHg x 133 Pa/mmHg = 9 283 Pa

    V - volume - 285 x 10⁻⁶ m³

    n - number of moles

    R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

    T - temperature in Kelvin - 25 °C + 273 = 298 K

    substituting these values in the equation

    9283 Pa x 285 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 298 K

    n = 1.067 x 10⁻³ mol

    decomposition of metal carbonate is as follows

    MCO₃ - - - > MO + CO₂

    stoichiometry of MCO₃ to CO₂ is 1:1

    therefore number of moles of MCO₃ heated = number of CO₂ moles formed

    number of MCO₃ moles = 1.067 x 10⁻³ mol

    molar mass = mass / number of moles

    molar mass = 0.158 g / 1.067 x 10⁻³ mol = 148 g/mol

    since carbonate molar mass is known -

    (molar mass of C x 1 C atom) + (molar mass of O x 3 O atoms)

    12 + 16x 3 = 12 + 48 = 60

    then mass of metal M - 148 - 60 = 88

    group II metal with molar mass of 88 is Ra - Radium
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