$\rm a)$ Phản ứng xảy ra:
$\rm\quad\quad\;\;\; Zn\:+\;HCl\to ZnCl_2+H_2\\(mol)\;\:0,1\to0,2\hspace{2,55cm}0,1$
$\rm b)$ Ta có: $\rm n_{Zn}=\dfrac{6,5}{65}=0,1\ (mol)$
$\rm ⇒V_{H_2}=0,1×22,4=2,24\ (lit)$
$\rm c)$ Từ phản ứng: $\rm n_{HCl}=0,2\ (mol)$
$\rm ⇒C\%_{HCl}=\dfrac{0,2×36,5}{200}×100\%=3,65\%$