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30 March, 13:35

elf-contained self-rescue breathing devices convert CO2 into O2 according to the following reaction: 4KO2 (s) + 2CO2 (g) → 2K2CO3 (s) + 3O2 (g) How many grams of KO2 are needed to produce 100.0 L of O2 at 20.0 °C and 1.00 atm?

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  1. 30 March, 13:39
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    394.05g of KO2

    Explanation:

    First, let us calculate the number of moles of O2 produced. This is illustrated below:

    V = 100L

    P = 1atm

    T = 20°C = 20 + 273 = 293K

    R = 0.082atm. L/K / mol

    n = ?

    PV = nRT

    n = PV / RT

    n = (1 x 100) / (0.082 x 293)

    n = 4.16moles of O2

    4KO2 + 2CO2 → 2K2CO3 + 3O2

    From the equation,

    4moles of KO2 produced 3moles of O2.

    Therefore, xmol of KO2 will produce 4.16moles of O2 i. e

    xmol of KO2 = (4.16 x 4) / 3 = 5.55moles.

    Converting 5.55moles O KO2 to grams, we have:

    Molar Mass of KO2 = 39 + (2x16) = 39 + 32 = 71g/mol

    Mass of KO2 = 5.55 x 71 = 394.05g

    Therefore, 394.05g of KO2 is needed for the reaction
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