Why can't we prove consistency of ZFC like we can for PA? this might be a silly question, but I was wondering: PA cannot prove its consistency by the incompleteness theorems, but we can "step outside" and exhibit a model of it, namely $\mathbb{N}$, so we know that PA is consistent.
Why can't we do this with ZFC? I have seen things like "if $\kappa$ is [some large cardinal] then $V_{\kappa}$ models ZFC", but these stem from an "if". 
Is this a case of us not having been able to do this yet, or is there a good reason why it is simply not possible?

Of course we can. And we do just that.
To "step outside of $\sf PA$" means that you assume the consistency of a far stronger theory, e.g. $\sf ZFC$, which lets you construct $\Bbb N$ as an object and prove it satisfies $\sf PA$.
When you say "assume that $\kappa$ is a cardinal such that $V_\kappa$ is a model of $\sf ZFC$" you effectively saying "We are stepping outside of $\sf ZFC$ into a theory "$\sf ZFC+\varphi$ for a suitable axiom $\varphi$, and there we can can find a model of $\sf ZFC$.
In fact, when you assume something like $V_\kappa$ is a model of $\sf ZFC$, you can even find a fairly canonical model of $\sf ZFC$: $L_\alpha$ for the least $\alpha$ satisfying $\sf ZFC$, where $L_\alpha$ is the $\alpha$th step in the constructible hierarchy. As to why it exists, that's another question. But it is there.
