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Homework Statement:

The following equilibrium reaction is known as the 'water gas shift' reaction:
$$CO(g) + H_2O(g) \Leftrightarrow H_2(g) + CO_2(g)$$
When initially 0.40mol of CO(g) and 1.00 mol of $$H_2O(g)$$ are present in a vessel at $$1200°C$$, the pressure is 2 atm. The amount of $$CO_2(g)$$ is 0.225 mol at equilibrium.
a) Calculate the total equilibrium amount, n, of the gas mixture. Hint: Use the ICE table
b) Calculate the extent of dissociation, α.
c) Calculate the number of moles of each gas, except the carbon dioxide gas in the mixture at equilibrium.
Relevant Equations:
 pV=nRT
This is how I started my ICE table:
I am not sure if my table is correct. When I work out the total equilibrium amount, n, using the 1.4α I get: $$n = 0.40  0.40α +1.00  1.00α +1.4α +0.225$$
and that gives $$n = 1.625 mol$$
Rounding to the correct significant figures: n = 1.6 mol
I then used the 1.6 mol to get the extent of dissociation, α, but my answer does not make any sense when I substitute it back into the 'n' equation.
Can anyone give me hint to help me to see where I am making a mistake?
Thank you very much.
$$CO(g)$$  $$H_2O(g)$$  $$H_2(g)$$  $$CO_2(g)$$  
Initial (mol)  0.40  1.00  0  0 
Change (mol)  0.40α  1.00α  1.4α  1.4α 
Equilibrium (mol)  0.400.40α  1.001.00α  1.4α  1.4α (or should I use 0.225) 
and that gives $$n = 1.625 mol$$
Rounding to the correct significant figures: n = 1.6 mol
I then used the 1.6 mol to get the extent of dissociation, α, but my answer does not make any sense when I substitute it back into the 'n' equation.
Can anyone give me hint to help me to see where I am making a mistake?
Thank you very much.