The problem is that ∃xp(x) is true, and ∃xq(x) is true means that for some x=x1, p(x) is true, and for some x=x2, q(x) is true. There is no guarantee that x1=x2, although it is not impossible.
However, ∃x(p(x) ∧ q(x)) means that for the same x=x0, p(x0) is true, and q(x0) is true. This cannot be implied from the first proposition, therefore
∃xp(x) ∧ ∃xq(x) => ∃x(p(x) ∧ q(x)) is a false statement.
