Is it possible that only a finite number of axioms among axiom schemata of ZFC is consistent? I was reading a wonderful argument of Hamkins in https://mathoverflow.net/a/51786 , saying that even a model $M$ of ZFC where $\neg CON(ZFC)$ holds has a "pseudo-model" $m$ within it which satisfies all axioms of ZFC, because $M$ has more axioms (coming from axiom schemata) than the axioms of ZFC.
Is it possible, then, that our metatheory is itself in this situation, in the sense that the ZFC (as applied to our metatheory) in itself inconsistent, but only a (huge but) finite number of axioms (coming from axiom schemata) of ZFC is consistent?
 A: No, it's not possible (at least as long as Zermelo set theory or some other similar base theory is consistent).
But unfortunately it's not for a very interesting reason: There are infinitely many formulas $\phi$ for which the corresponding instance of the replacement axiom schema
$$\forall u \forall p_1, \dots, p_n \big[ (\forall x\in u)(
\exists ! y\; \phi(x, u, p_1, \dots, p_n)\Rightarrow \dots \big]$$
is vacuously true because the antecedent in the implication is provably false for all non-empty $u$ and for all values of $p_1,\dots,p_n.$  (For example, one could take $\phi$ to be $x\ne x,\; x \ne x\wedge x \ne x,\; x \ne x\wedge x\ne x\wedge x\ne x,\;$ etc.)  Excluding $u=\emptyset$ here is fine since in that case the axiom schema is trivially true anyway.
So $V_{\omega+\omega}$ is a model of ZC plus infinitely many instances of the replacement axiom.

A more interesting similar question might be: Is there a recursively enumerable axiomatization $T$ in the language of set theory which is provably equivalent to ZFC but in which there are no redundant axioms? (A redundant axiom would be one which can be proven from the other axioms of $T$.)  (For this question, assume the consistency of ZFC so that the question has some substance.)
For this revised question, I think we can change the replacement axiom schema, as follows.  For each $m\lt\omega,$ let $\sigma_m$ be universal for $\Sigma_m$ formulas with three free variables.  Then, for each $m\lt\omega,$ we have the single axiom
$$(\forall k\lt\omega)\big(\text{traditional replacement axiom schema for  }\phi(x,u,p) = \sigma_m(k, x,u,p)\big).$$
I haven't checked through all the details here; in particular, you'd need to be sure that you can prove the traditional replacement axiom instances where $\phi$ has more parameters, but that should work by coding $p_1, \dots, p_n$ in a single parameter via Kuratowski pairing.  (You’d also need to be sure that you include enough axioms to prove the necessary properties of $\Sigma_m\text{-universal}$ sets, etc.)
If this all goes as expected, then this revised (but equivalent) version of ZFC would satisfy OP's requested property: it's possible that only finitely many of these new axioms are consistent.
