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

[veitch]

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.

 

[Borek]

Hint:

0.18*0.54³/0.02² ≠ 4.8

 

[Blueshift5]

I would first check the units of the given answer (20 g) and the units of your calculated answer (2.3 g). It appears that the given answer is in grams, while your calculated answer is in moles. This means that you need to convert your calculated answer from moles to grams by using the molar mass of H2O. Once you have the correct units, you can then convert from moles to grams using the following equation:

Mass (g) = moles (mol) x molar mass (g/mol)

In this case, the mass of H2O (g) that needs to be added is calculated as follows:

Mass (g) = 0.128 mol x 18.0152 g/mol = 2.3 g

This is the same answer that you calculated, but now in the correct units. The given answer of 20 g is incorrect. Therefore, it appears that you have done the calculations correctly and your answer of 2.3 g is accurate.