Are most Kähler manifolds non-projective? Since now-a-days lots of research activities are happening to prove many results for compact Kähler manifolds which are known for projective varieties, I was wondering are there plenty of non-projective Kähler manifolds? If yes, where can I find some explicit examples? I am aware of the theorem that a generic complex torus $\mathbb{C}^g/\Lambda$ is non-projective.
 A: Expanding a bit on Francesco Polizzi's remark, complex K3 surfaces are Kähler [1], the moduli space $M$ of complex K3 surfaces is an irreducible 20-dimensional complex variety, and within $M$, the set of K3 surfaces that are algebraic varieties is a countable union of disjoint subvarieties $F_g$ for integers $g\ge2$. Here $F_g$ has dimension 19, and the K3 surfaces in $F_g$ are those that have a primitive line bundle $\mathcal L$ satisfying $c_1(\mathcal L)^2=2g-2$.
So this is the analogue for K3 surfaces of the example you know for complex tori, i.e., in the moduli space of complex tori of a given dimension, the projective ones form a countable union of proper subvarieties.
[1] Siu, Y. T. (1983), Every K3 surface is Kähler, Inventiones Mathematicae 73 (1): 139–150
A: See Claire Voisin's amazing results on the subject, or the published version: 
On the homotopy types of compact kaehler and complex projective manifolds,
Inventiones Math. 157 2 (2004), 329 - 343. 
(ArXiv) Abstract: We show that in every dimension greater than or equal to 4, there exist compact Kaehler manifolds which do not have the homotopy type of projective complex manifolds. Thus they a fortiori are not deformation equivalent to a projective manifold, which solves negatively Kodaira's problem.
We give both non simply connected (of dimension at least 4) and simply connected (of dimension at least 6) such examples. 
A: I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't find any in complex dimension one, so let's look in  dimension two. By classification of surfaces, the non algebraic surfaces would have Kodaira dimensions $\kappa=0$ or $1$. The $\kappa=0$ cases are, I believe, necessarily tori or K3's (there are some other examples, but these are either algebraic or non Kähler). As already noted by you and Francesco, the algebraic surfaces form a proper subset in moduli for these cases.
The $\kappa=1$ surfaces are elliptic surfaces, and I expect that there should be plenty of non algebraic examples, although I don't have an example on hand.
Added later For my own reasons, I thought about this longer than I normally would. Here's  an explicit example. Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma=\pi_1(C)$, and $\tilde C$ the universal cover. Choose an   elliptic curve $E$ and a group homomorphism $h:\Gamma\to E$. Define an action of $\Gamma$ on $\tilde C\times E$ by $\gamma(x,y)= (\gamma x, y+h(\gamma))$, and let $S$ be the quotient. 

$S$ is Kähler. If $h$ has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde C\times E$ has an invariant Kähler metric. 
For the second statement, assume that $h$ has infinite image.
Projection on the first factor gives a holomorphic map $f:S\to C$.  The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$  which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.
Additional remarks: This has $\kappa=1$ when $g>1$. When $g=1$, one can see, with a bit of thought, that $S$ is torus which contains an elliptic curve but it is not isogenous to a product (i.e. Poincar\'e reducibility fails for tori). 
