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Homework Help: What mass must be added to raise concentration? Am I doing this right?

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    CH4 (g) + H2O (g) <- -> CO (g) + 3 H2 (g)
    .2 M CH4 and .2 M H2O are added to an empty 4 L sealed vessel at 1400 K. When equilibrium is established, the concentration of H2 is .444 M. Calculate Kc... which I did and got 4.8 which is the given answer.

    The real problem I have is part b) What mass of H2O (g) must be added to the equilibrium mixture in order to raise the concentration of H2 to .54 mol/L? The answer is supposed to be "20. g"

    3. The attempt at a solution

    using the equilibrium concentrations I calculated for the first part as initial concentrations

    CH4 H2O CO 3H2
    I .052 .052 .148 .444
    C -.032 -.032 +.032 +.096
    E .02 .02 .18 .54

    So to change the .444 M H2 to .53 M... the reactants would need a third more added to them? and .032 M/L in 4.0 L would be 0.128 mol... if the molar mass of H2O is 18.0152 then that would be 2.3 g... which is nowhere near 20 g

    Am I way off? I'm confused.
    Last edited: Mar 12, 2009
  2. jcsd
  3. Mar 13, 2009 #2


    User Avatar

    Staff: Mentor


    0.18*0.543/0.022 ≠ 4.8
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