Does a long exact sequence of flat modules remain exact after tensoring with an arbitrary module? In Liu's Algebraic Geometry and Arithmetic Curves, Proposition 1.2.6 states that given any short exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ with $M''$ flat, taking the tensor product of this with any module $N$ gives an exact sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$.

Later in the book, in Exercise 5.2.8, page 193, he suggests using this Proposition to show that given a long exact sequence which is eventually $0$:

$$ 0 \rightarrow M_0 \rightarrow M_1 \rightarrow \cdots $$

of flat modules, taking the tensor product with an arbitrary module still gives an exact sequence 

$$ 0 \rightarrow M_0 \otimes N \rightarrow M_1 \otimes N \rightarrow \cdots$$

How does this generalization follow? It appears that in the proof of the Proposition, he doesn't really use the surjectivity of $M \rightarrow M''$, but then again, all he proves is that $M' \otimes N \rightarrow M'' \otimes N$ is injective; the full exactness being something that always happens for tensor products of short exact sequences.
In fact, Liu says (on page 193, Exercise 2.8(b)) that the sequence is "zero from a finite rank on". Then you start the breaking from the rightmost side of the sequence: $0\to X_{n-1}\to M_{n-1}\to M_n\to 0$, and so on. You know that $M_{n-1}$ and $M_n$ are flat, so $X_{n-1}$ is also flat, and so on.
let $$0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow \cdots$$ be exact. 
then there is exact sequences (break):
$$0 \rightarrow M_0 \rightarrow M_1 \rightarrow C_1 \to 0$$ and $$0\to C_1 \rightarrow M_2 \rightarrow C_2 \to 0$$  and so on...  

now tensor with $N$ and join sequences again.