Total entropy change for a spontaneous reaction For the oxidation of iron,
$$\ce{4 Fe(s) + 3 H2O(g) -&gt; 2 Fe2O3(s) + 3 H2(g)}$$
the entropy change is $ \pu{–549.4 J K^{-1} mol^{-1}}$ at $\pu{298 K}.$ In spite of the negative entropy change of this reaction, why is the reaction spontaneous? $(\Delta_\mathrm{r}H^\circ$ for this reaction is $\pu{–1648 kJ mol^{–1}})$

In this question the solution given is that we consider the total entropy change for the reaction, i.e. the entropy change for a system plus the entropy change of surroundings, and that comes out to be positive.
But shouldn't the entropy change for surroundings (calculated by $q_\mathrm{rev}/T)$ be the negative of the entropy change for the system (heat absorbed by the surroundings is same as the heat emitted by system)? Thus, shouldn't the total entropy change be zero?
The initial state of your system consists of 4 moles of $\ce{Fe(s)}$ and 3 moles of water vapor at $\pu{298 K}$ and $\ce{1 bar}$ in separate containers. The final state of your system consists of 2 moles of $\ce{Fe2O3(s)}$ and 3 moles of $\ce{H2(g)}$ at $\pu{298 K}$ and $\pu{1 bar}$ in separate containers.
The $\Delta H^\circ$ and $\Delta S^\circ$ correspond to a process for taking this system reversibly from the initial state to the final state. They don't include the changes in enthalpy and entropy of the surroundings in executing this process.
Do you have an idea of what such a process would be?