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

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SUMMARY

The discussion centers on calculating the mass of H2O required to increase the concentration of H2 in a chemical equilibrium involving the reaction CH4 (g) + H2O (g) <-> CO (g) + 3 H2 (g). The initial concentrations were 0.2 M for CH4 and H2O, and upon reaching equilibrium, the concentration of H2 was found to be 0.444 M. The user calculated the equilibrium concentrations and attempted to determine that 20 g of H2O must be added to achieve a concentration of 0.54 M for H2, but their calculations yielded a different result, indicating a misunderstanding of the equilibrium shift and mass calculations.

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veitch
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Homework Statement



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"

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:
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Hint:

0.18*0.543/0.022 ≠ 4.8
 

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