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26 April, 04:24

What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.10 atm?

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  1. 26 April, 04:42
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    When the reaction equation is:

    CaSO3 (s) → CaO (s) + SO2 (g)

    we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

    to get the moles of SO2 we are going to use the ideal gas equation:

    PV = nRT

    when P is the pressure = 1.1 atm

    and V is the volume = 14.5 L

    n is the moles' number (which we need to calculate)

    R ideal gas constant = 0.0821

    and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

    so, by substitution:

    1.1 * 14.5 L = n * 0.0821 * 285.5

    ∴ n = 1.1 * 14.5 / (0.0821*285.5)

    = 0.68 moles SO2

    ∴ moles CaSO3 = 0.68 moles

    so we can easily get the mass of CaSO3:

    when mass = moles * molar mass

    and we know that the molar mass of CaSO3 = 40 + 32 + 16 * 3 = 120 g/mol

    ∴ mass = 0.68 moles * 120 g/mol = 81.6 g
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