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7 May, 10:51

An equilibrium mixture contains 0.350 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00 L container. CO (g) + H 2 O (g) - ⇀ ↽ - CO 2 (g) + H 2 (g) How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

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  1. 7 May, 12:09
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    0.8524 moles have to be added

    Explanation:

    Step 1: Data given

    An equilibrium mixture contains 0.350 mol of each of the products

    Each product contains 0.200 moles

    Volume = 1.00L

    Step 2: The balanced equation

    CO (g) + H2O (g) ⇆ CO2 (g) + H2 (g)

    Step 3: Calculate Kc

    Kc = [CO2][H2] / [CO][H2O]

    Kc = [0.350][0.350]/[0.200][0.200]

    Kc = 3.0625

    Step 4: The initial number of moles

    nCO2 = 0.200 moles = 0.200 M

    nCO = 0.200 moles = 0.200 M

    nCO2 = O. 350 + n moles = 0.350 + n M

    nH2 = 0.350 M

    Step 5: the number of moles at the equilibrium

    nCO2 = 0.200 moles + X = 0.300 mol

    ⇒ X = 0.100 moles

    nCO = 0.200 moles + 0.100 moles = 0.300 moles

    nCO2 = 0.350 + n - 0.100 moles = 0.250 + n

    nH2 = 0.350 - 0.100 = 0.250 moles

    Step 6: Calculate Kc

    Kc = [CO2][H2] / [CO][H2O]

    3.0625 = (0.250 + n) (0.250) / (0.300) (0.300)

    0.2756 = (0.250 + n) (0.250)

    1.1024 = 0.250+n

    n = 0.8524 moles

    0.8524 moles have to be added
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