Suppose $X$ is a non-singular projective rational surface over an algebraically closed field and $C \subset X$ is a non-singular irreducible rational curve on $X$ such that the complete linear system $|C|$ is positive dimensional.
Does it follow that the general member $|C|$ is nonsingular and irreducible, or at least irreducible?
My initial idea to deal with this was using Bertini Theorem, which claims that if $|C|$ does not have fixed components and the dimension of the rational map $$ \Phi: X \dashrightarrow \mathbb{P}(H^0(X, \mathcal{L}(C))) $$ induced by $|C|$ is greater or equal than $2$, then the general member is irreducible. So instead one could ask
Does $C$ irreducible and non-singular imply $\dim \Phi(X) \ge 2$?
Just to expand my comment. Blow up the finitely many (possibly) points in the base locus of $|C|$, to get a morphism $f:X\to Y$ with a fiber precisely $C$. Then, either use Bertini (or Stein factorization) to show that general fiber is irreducible. Then show that these are general elements of $|C|$.