A mixture of CO(g) and H2O(g) at equilibrium contains 0.6 mol CO, 0.6 mol H2O, 0.1 mol CO2, and 0.1 mol H2 in a 2 dm3 flask. Kc = 4. Calculate the mass of CO at equilibrium.

The balanced equation is:

$$\text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)}$$

Initial moles:
$\text{n(CO)} = 0.6 \text{ mol}$
$\text{n(H}_2\text{O)} = 0.6 \text{ mol}$
$\text{n(CO}_2\text{)} = 0.1 \text{ mol}$
$\text{n(H}_2\text{)} = 0.1 \text{ mol}$
Volume of flask, $\text{V} = 2 \text{ dm}^3$
Equilibrium constant, $\text{K}_c = 4$

Let $x$ be the change in moles for the reaction.

$$\begin{array}{|l|c|c|c|c|} \hline & \text{CO(g)} & \text{H}_2\text{O(g)} & \text{CO}_2\text{(g)} & \text{H}_2\text{(g)} \\ \hline \text{Initial (mol)} & 0.6 & 0.6 & 0.1 & 0.1 \\ \text{Change (mol)} & -x & -x & +x & +x \\ \text{Equilibrium (mol)} & 0.6-x & 0.6-x & 0.1+x & 0.1+x \\ \hline \end{array}$$

The equilibrium concentrations are calculated by dividing the equilibrium moles by the volume ($2 \text{ dm}^3$).

$$K_c = \frac{\left(\frac{0.1+x}{2}\right)\left(\frac{0.1+x}{2}\right)}{\left(\frac{0.6-x}{2}\right)\left(\frac{0.6-x}{2}\right)}$$ $$4 = \frac{\frac{(0.1+x)^2}{4}}{\frac{(0.6-x)^2}{4}}$$ $$4 = \frac{(0.1+x)^2}{(0.6-x)^2}$$

Take the square root of both sides:

$$\sqrt{4} = \sqrt{\frac{(0.1+x)^2}{(0.6-x)^2}}$$ $$2 = \frac{0.1+x}{0.6-x}$$

Now, solve for $x$:

$$2(0.6-x) = 0.1+x$$ $$1.2 - 2x = 0.1 + x$$ $$1.2 - 0.1 = x + 2x$$ $$1.1 = 3x$$ $$x = \frac{1.1}{3} \text{ mol}$$ $$x \approx 0.3667 \text{ mol}$$

$$\text{n(CO)}_{\text{eq}} = 0.6 - x$$ $$\text{n(CO)}_{\text{eq}} = 0.6 - \frac{1.1}{3}$$ $$\text{n(CO)}_{\text{eq}} = \frac{1.8}{3} - \frac{1.1}{3}$$ $$\text{n(CO)}_{\text{eq}} = \frac{0.7}{3} \text{ mol}$$ $$\text{n(CO)}_{\text{eq}} \approx 0.2333 \text{ mol}$$

Using atomic masses: $\text{C} = 12.01 \text{ g/mol}$, $\text{O} = 16.00 \text{ g/mol}$

$$\text{M(CO)} = 12.01 + 16.00 = 28.01 \text{ g/mol}$$

$$\text{m(CO)} = \text{n(CO)}_{\text{eq}} \times \text{M(CO)}$$ $$\text{m(CO)} = \frac{0.7}{3} \text{ mol} \times 28.01 \text{ g/mol}$$ $$\text{m(CO)} = \frac{19.607}{3} \text{ g}$$ $$\text{m(CO)} \approx 6.5356 \text{ g}$$

Rounding to two decimal places:

$$\text{m(CO)} = \boxed{\text{6.54 g}}$$