Given that $X_{n}$ converges in probability to $2$.
Also if $Y_{n}$ converges in distribution to $N(0,5)$.
Then what is the asymptotic distribution of $Y_{n} X_{n}$?
If $X_{n} \overset{p}{\rightarrow} 2$ then this implies that $X_{n} \overset{d}{\rightarrow} 2$
And then if $Y_{n} \overset{d}{\rightarrow} N(0,5)$, then does this mean that $$Y_{n} X_{n} \overset{d}{\rightarrow} 2N(0,5)=N(0,10)?$$
$g(x) := f(2,x)$ is a continuous bounded funtion for every continuous bounded function $f:\Bbb R^2 \to \Bbb R $. Thus due to $Y_n \to Z \sim \mathcal N (0,5)$ in distribution $$\Bbb E[f(2,Y_n)] = \Bbb E [g(Y_n)] \to \Bbb E[g(Z)] = \Bbb E[ f(2,Z)]$$ Thus $(2,Y_n) \to (2,Z)$ in distribution.
Furthermore for any $\varepsilon > 0 $ we have $$\Bbb P ( \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon) = \Bbb P (\vert X_n - 2\vert > \varepsilon) \to 0$$ For a continuous $f$ with compact support (thus uniformly continuous) and $\delta , \varepsilon > 0$ such that $\Vert (x,y) - (u,v)\Vert \le \varepsilon \Rightarrow \vert f(x,y) -f(u,v)\vert < \delta$ $$\vert\Bbb E [f(X_n , Y_n)] - \Bbb E[ f(2,Z)] \vert \\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + \vert \Bbb E[f(X_n , Y_n )] - \Bbb E[f(2, Y_n)] \vert\\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + \Bbb E[\vert f(X_n , Y_n )-f(2, Y_n) \vert ; \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon]+ \Bbb E[\vert f(X_n , Y_n )-f(2, Y_n) \vert ; \Vert (X_n , Y_n) - (2, Y_n) \Vert \le \varepsilon]\\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + 2\Vert f\Vert_{\infty}\Bbb P( \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon)+ \delta $$
All in all $$\limsup_{n\to\infty} \vert\Bbb E [f(X_n , Y_n)] - \Bbb E[ f(2,Z)] \vert \le \delta \to 0 $$, as $\delta\to 0$. Thus $(X_n , Y_n) \to (2, \mathcal N(0,5))$ in distribution by the portmanteau theorem.
Now for a continuous bounded function $f:\Bbb R \to \Bbb R$, the function $g(x,y) := f(xy)$ is continuous and bounded. Therefore $$\Bbb E[f(X_n Y_n)] \to \Bbb E[f(2Z)]$$ and thus $X_nY_n\to 2Z \sim \mathcal N (0, 20)$ in distribution
By the way: In my notation $Z\sim \mathcal N(0,5)$ means $\Bbb E[Z^2] = 5$