Carbon monoxide reacts with steam in a closed vessel: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)…

(a) The quadratic. At equilibrium: CO $(0.600 - x)$, steam $(0.900 - x)$, $\text{CO}_2$ $x$ and $\text{H}_2$ $x$, all in $\text{mol}$. Each concentration is the amount divided by $3.00\ \text{L}$, and the volumes cancel because $\Delta n_g = 0$:

This gives $0.44x^2 - 2.160x + 0.7776 = 0$. The root that is smaller than $0.600$ is $x = 0.391\ \text{mol}$ of hydrogen.

(b) Percentage. $\dfrac{0.391}{0.600} \times 100 = 65.2\%$ of the CO has reacted.

(c) Concentration. $[\text{CO}] = \dfrac{0.600 - 0.391}{3.00}\ \text{mol L}^{-1} = 0.0696\ \text{mol L}^{-1}$.

(d) $K_p$. $\Delta n_g = 2 - 2 = 0$, so $K_p = K_c(RT)^0 = 1.44$, with no unit.

(e) More steam. Just after the steam goes in, $Q_c = \dfrac{0.391^2}{0.209 \times 1.009} = 0.726$, less than $K_c$, so more CO reacts. The final mixture is the one you would get from $0.600\ \text{mol}$ of CO and $1.400\ \text{mol}$ of steam:

So $0.060\ \text{mol}$ more hydrogen forms, as Le Chatelier's principle predicts.