Question
When an amount ammonia is added at $\pu{600 K}$ in a $\pu{1 L}$ container the following reaction takes place:
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
The equilibrium constant $K_c = 4.20$ at $\pu{600 K}.$
At equilibrium, it is known that $\pu{0.200 mol}$ of $\ce{N2}$ gas exist in the container. What amount of ammonia was added at the start of the reaction? Choose from the answers below:
$\pu{0.826 mol};$ $\pu{0.482 mol};$ $\pu{1.226 mol};$ $\pu{0.400 mol};$ $\pu{0.800 mol}.$
My attempt
I created an ICE table, but I think it's wrong:
$$ \begin{array}{lccc} \ce{&N2(g) &+ &3 H2(g) &<=> &2 NH3(g)} \\ \text{I} & 0 && 0 && y \\ \text{C} & +x && +3x && -4x \\ \text{E} & 0.2 && 0.6 && y-0.8 \\ \end{array} $$
$$K_c = \frac{(y - 0.8)^2}{0.200\times 0.6^3}$$
Solving the equation with Maple gives me none of the answers above. I must have messed up on the table. Can anyone tell me where I messed up?