It is probably a trivial question, but I have no clue about the answer. I know that the elements of $\omega_1$, the first uncountable ordinal, are the countable ordinals, of which there are uncountably many, but I am not sure if that applies to the number of limit ordinals before $\omega_1$ too.
No: the set of countable limit ordinals has the same cardinality as the set of countable ordinals, i.e., $\omega_1$ (or $\aleph_1$, for those who prefer the aleph notation). Let $\Lambda$ be the set of non-successor countable ordinals (i.e., the countable limit ordinals together with $0$); then the map
is a bijection, where $\cdot$ is ordinal multiplication.
Here is a slightly worse proof than that given by Brian.
Suppose that the $A$ is the set of countable limit ordinals, then $\bigcup A=\delta=\sup A$ is a countable limit ordinal, therefore $\delta\in A$, but this means that $\delta=\max A$. It is easy to see that $\delta+\omega>\delta$ and it is a countable limit ordinal, in contradiction to the fact that $\delta$ is the largest countable limit ordinal.