The equilibrium constant for the following reaction at $\pu{600^{\circ}C}$ is 4.0. Initially, two moles of $\ce{CO}$ and one mole of $\ce{H2O}$ were mixed in a $\pu{1.0 L}$ container. Determine the concentration of all species at equilibrium.
Attempt #1:
$$\ce{CO (g) + H2O (g) -> CO2 (g) + H2 (g)}$$
\begin{array}{|r|c|c|c|} \hline \text{Initial}~(M) & 2.0 & 1.0 & 0 & 0 \\ \hline \text{Change}~(M) & -x & -x & +x & +x \\ \hline \text{Equilibrium}~(M) & 2.0 – x & 1.0 - x \\ \hline \end{array}
Plug $x$ into the equilibrium expression and solve for $x$. $4.0 = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.
Determine concentrations: the equilibrium values become 2.0 – 0.85, 1.0 - 0.85, 0.85, and 0.85, giving 1.2, 0.1, 0.85, and 0.85.
Where does the 0.85 come from? Could it be cross multiplied some way or is there another way this is done?
Furthermore, I want to make sure that I've done the 3.00 correctly for $\ce{Fe(SCN)2+}$.
Consider the reaction represented by the equation: $$\ce{Fe^3+ (aq) + SCN- (aq) -> Fe(SCN)2+ (aq)}$$
Attempt #2:
\begin{array}{ccc} \text{Initial} & \pu{6.00 M}\ \ce{Fe^{3+} (aq)} & \pu{6.00 M}\ \ce{SCN^{−} (aq)} \\ \text{Equilibrium} & \pu{? M}\ \ce{FeSCN2^+ (aq)}, \ K = 0.33 \end{array}
\begin{align} \frac{1}{3}x^2 - \frac{124}{25}x + \frac{297}{25} &= 0 & x_{(1;2)} &= \frac{\color{red}{\pmb{-}}(-\frac{124}{25})\pm\sqrt{\left(\frac{124}{25}\right)^2 - 4\cdot\frac13\frac{297}{25}}}{\frac{2}{3}} \\ && x_{(1;2)} &= \frac{124/25\pm\sqrt{\frac{15376}{(625)}-\frac{15601}{1000}}}{\frac{2}{3}} \\ && &= \left(124/25\pm{3}\right)\cdot\frac{3}{2} \\ && x_{(1;2)} &= 12;3 \end{align}
