$\ce{Cr2O3}$ can be reduced to the elements chromium with high temperature and carbon by this two reactions:
$$ \begin{align} \ce{Cr2O3(s) + 3C(s) &-> 2Cr(s) + 3 CO(g)} \label{rxn:1}\tag{1}\\ \ce{Cr2O3(s) + \frac{3}{2}C(s) &-> 2Cr(s) + \frac{3}{2}CO2(g)} \label{rxn:2}\tag{2} \end{align} $$
The following image shows the equilibrium constants.
Using the image, we can get $K_1$, $\Delta_rG^\circ_1$, $\Delta_rH^\circ_1$, $\Delta_rS^\circ_1$ of reaction $\eqref{rxn:1}$ as well as $K_2$ for $T = 1500\,\mbox{K}$.
Assume all of them are given. Now:
In a closed container with $V = 1.00\,\mbox{dm}^3$ inner volume, we put $5.00\,\mbox{g}$ of $\ce{Cr2O3}$ and $20.0\,\mbox{g}$ carbon. For $T = 1500\,\mbox{K}$ we have the reaction equilibrium $\eqref{rxn:1}$ and $\eqref{rxn:2}$.
Problem: What mass (in g) reduced chromium do we have in the equilibrium?
Question: Now I don't know how to solve the problem. I can do everything above like determining the Gibbs-Energy or the equilibrium constants, but what are my though now to get the amount of reduced chromium?
First, I don't get exactly what it means if they say "in the equilibrium". Does that refer to $\eqref{rxn:1}$ or $\eqref{rxn:2}$ or both?
Further, how do I actually calculate it?