Consider the embedding $$\iota: B(H_1) \odot B(H_2)\to B(H_1 \otimes H_2),$$ where $T\in B(H_1)$ and $S\in B(H_2)$ will be mapped to $T\otimes S\in B(H_1\otimes H_2)$ with $T\otimes S(v\otimes w)=T(s)\otimes S(w)$ (and $H_1$ and $H_2$ are complex Hilbert spaces).
$B(H_1) \odot B(H_2)$ denotes the tensorproduct of $B(H_1)$ and $B(H_2)$ as a $*$-algebra and $H_1 \otimes H_2$ is the tensor product auf complex Hilbert space.
Why is $\iota$ not surjective, in general?
For a counterexample it is necessary to take $H_1$ and $H_2$ infinite-dimensional, in the finite-dimensional case $\iota$ is always an isomorphism. If we take $H_1=H_2=\ell^2$, I'm searching for $F\in B(\ell^2\otimes \ell^2)$ such that $F\notin Im(\iota)$, but I don't know which operator $F$ will work.
An other idea which comes to my mind is to use Why is $\overline{B(l^2)\odot B(l^2)}^{\| \enspace \|_{op}}\neq B(l^2\otimes l^2)?$ but at the moment I'm not sure how to use it in detail.
If either $H_1$ or $H_2$ is finite-dimensional, then $\iota$ is surjective. Otherwise, the domain of $\iota$ is incomplete (being the algebraic tensor product of two infinite-dimensional objects) but the codomain is complete, so they cannot be isomorphic even as normed spaces.
However, even after completion there is no isomorphism. If $H_1, H_2$ are separable, then you may identify them with $\ell_2$, and use the linked solution (it can be extended to the non-separable case too). It turns out that these objects are not even isomorphic as Banach spaces (see this note).