Đáp án: $C$
Giải thích các bước giải:
$n_{FeCl_3}=\dfrac{32,5}{162,5}=0,2(mol)$
$2Fe+3Cl_2\buildrel{{t^o}}\over\to 2FeCl_3$
$\Rightarrow n_{Cl_2}=0,3(mol)$
$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O$
$\Rightarrow n_{HCl}=\dfrac{16}{5}n_{Cl_2}=0,96(mol)$
$m_{dd HCl}=0,96.36,5:37\%=94,702g$
$\to V_{HCl}=\dfrac{94,702}{1,19}=79,58ml$