Determining equilibrium concentrations from initial conditions and equilibrium constant


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> 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.
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### 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}



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> Plug $x$ into the equilibrium expression and solve for $x$. $4.0 = 2[x]$, so $x = 0.85 [1.0 – x][2.0 – x]$.
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> 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.
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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: $$\ce{Fe^3+ (aq) + SCN- (aq) -> Fe(SCN)2+ (aq)}$$
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### 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}




The text



> 
> 4.0 = [x]2 so x = 0.85 [1.0 – x][2.0 – x] 
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appears to be an error. The proper equation is:


$$4 = \frac{x^2}{(2-x)(1-x)}$$


The 0.85 is one of two roots of the quadratic equation you'll solve, using


$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$


after you simplify the first expression, above.


As for the rest of your question, please expand: What does



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> the 3.00
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mean?


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