Is the exponential map for $\text{Sp}(2n,{\mathbb R})$ surjective? For $\mathfrak{g} := {\mathfrak s}{\mathfrak p}(2n,\mathbb{R})$ and $G = \text{Sp}(2n,{\mathbb R})$, is the exponential map
\begin{equation}
\text{exp} : \mathfrak{g} \to G
\end{equation}
surjective?
If not then is it possible to access all of the symplectic group $\text{Sp}(2n,\mathbb{R})$ via a product, i.e. $e^Xe^Y$, such that $X,Y \in {\mathfrak s}{\mathfrak p}(2n,\mathbb{R})$? If so what is the minimum number of products needed?

In his comment, user148177 already explained that the exponential function of $\text{Sp}_{2n}({\mathbb R})$ is not surjective. 
Two factors, however, suffice: 

Theorem (Polar decomposition) Any $g\in\text{Sp}_{2n}({\mathbb R})$ can uniquely be written as a product $g = h\cdot\text{exp}(X)$ with $h\in SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$ and $X\in\text{Sym}_{2n}({\mathbb R})\cap{{\mathfrak s}{\mathfrak p}}_{2n}({\mathbb R})$. 

Reference: Hilgert-Neeb: The structure and geometry of Lie groups, Proposition 4.3.3.
As a compact Lie group $SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$ has surjective exponential, so $\text{Sp}_{2n}({\mathbb R})=\text{exp}({\mathfrak s}{\mathfrak p}_{2n}({\mathbb R}))^2$.
