Why can we not always take a Kähler class to be in rational cohomology? Given a Kähler manifold $(X,\omega)$ we know that its Kähler class lies in an open cone of $H^{1,1}(X) \cap H^2 (X,\mathbb{R})$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ we should be able to find a Kähler class belonging to $H^{1,1}(X) \cap H^2 (X,\mathbb{Q})$. 
Such a class would be represented by a closed positive form of type (1,1), therefore X would be projective thanks to Kodaira theorem.
I know this reasoning has to be wrong since not every Kähler manifold is projective. Can you tell me where?
Thanks.
 A: It might happen that $H^{1,1}(X)\cap H^2(X,\mathbb{Q})=0$ while $H^{1,1}(X)\cap H^2(X,\mathbb{R})\ne 0$. This is the case e.g. when $X=\mathbb{C}^n/\Lambda$, where $\Lambda$ is a generic lattice. 
Nevertheless your argument can be easily applied to prove that a compact Kahler manifold is algebraic provided $H^{2,0}(X)=0$.
A: Since Artie Prendergast-Smith is not expanding his comment in an answer, let me do it. As I said in the comments, his comment is essentially THE answer to the OP question. But let me give some more details.
So, let $V_\mathbb Q$ be a finite dimensional $\mathbb Q$-vector space, and let $V_\mathbb R:=V_\mathbb Q\otimes_\mathbb Q\mathbb R$ and $V_\mathbb C:=V_\mathbb Q\otimes_\mathbb Q\mathbb C$ (at the end of the day, thanks to the universal coefficient theorem, you have to think at $V_\mathbb K$, for $\mathbb K=\mathbb Q, \mathbb R, \mathbb C$, as $H^2(X,\mathbb K)$).
Then, on $V_\mathbb C=V_\mathbb R\oplus i V_\mathbb R$, you have a natural conjugation and $V_\mathbb R=\{v\in V_\mathbb C\mid v=\bar v\}$.
Now, suppose you have a decomposition (a so-called pure Hodge structure of weight $2$) on $V_\mathbb C$:
$$
V_\mathbb C=V^{2,0}\oplus V^{1,1}\oplus V^{0,2},
$$
where $V^{p,q}=\overline{V^{q,p}}$. Finally, call 
$$
V^{1,1}_\mathbb R=V_\mathbb R\cap V^{1,1}=\{v\in V^{1,1}\mid v=\bar v\}.
$$
Observe, first of all, that, as in the last sentence of semyon alesker's answer, if $V^{2,0}=\{0\}$, then $V_\mathbb C=V^{1,1}$, and thus 
$$
V^{1,1}_\mathbb R=V_\mathbb R.
$$
In this case, your reasoning works, since $V_\mathbb Q$ is dense in $V_\mathbb R=V^{1,1}_\mathbb R$ and therefore any open set in $V^{1,1}_\mathbb R$ (in particular any open cone...) contains some ''rational'' point.
In the general case, everything depends on the relative positions of $V_\mathbb R$ and $V^{2,0}$, $V^{0,2}$, and $V^{1,1}$. Now, the hypothesis of $X$ being Kähler can be translated in $V^{1,1}_\mathbb R\ne\{0\}$, so that in fact $V_\mathbb R$ always intersects in a non trivial way $V^{1,1}$, but it principle it may very well happen (and it happens, indeed, cf. the first part of the answer of semyon alesker), that $V_\mathbb R\not\subset V^{1,1}$ so that $\dim V^{1,1}_\mathbb R<\dim V_\mathbb R$. 
In this case an open set in $V^{1,1}_\mathbb R$, is just an open set (for the induced topology) in a (real) vector subspace of $V_\mathbb R$, and not in the whole $V_\mathbb R$! Now, of course a dense subset of $V_\mathbb R$ (like $V_\mathbb Q$ is, but now I am speaking in general) needs not to be dense in all subsets of $V_\mathbb R$; even worst it may very well have empty intersection with another subset of $V_\mathbb R$. In the case of $V_\mathbb Q$, it is dense and always contains the zero vector, so the worst case is that it intersects $V^{1,1}_\mathbb R$ only in zero (and this is the case for a generic torus!). 
And here comes the comment of Artie Prendergast-Smith. Just think at $V_\mathbb Q=\mathbb Q^3$ endowed with the following decomposition:
$$
V_\mathbb C=V^{2,0}\oplus V^{1,1}\oplus V^{0,2},
$$
where $V^{2,0}=\operatorname{Span}_\mathbb C(e_1+ie_3)$, $V^{1,1}=\operatorname{Span}_\mathbb C(e_1+\sqrt 2 e_2)$ and $V^{0,2}=\operatorname{Span}_\mathbb C(e_1-ie_3)$ (here $\{e_1,e_2,e_3\}$ is the canonical basis of $V_\mathbb K$). Then, $V^{p,q}=\overline{V^{q,p}}$ and $V^{1,1}_\mathbb R=\operatorname{Span}_\mathbb R(e_1+\sqrt 2 e_2)$. Then, no non-zero vector in $V^{1,1}_\mathbb R$ lies in $\mathbb Q^3$.
