a,
$n_{HCl}=4(mol)$
$MnO_2+4HCl\to MnCl_2+Cl_2+2H_2O$
$\Rightarrow n_{Cl_2}=1(mol)$
$V_{Cl_2}=1.22,4=22,4l$
$K_2Cr_2O_7+14HCl\to 2KCl+2CrCl_3+3Cl_2+7H_2O$
$\Rightarrow n_{Cl_2}=\dfrac{3}{14}.4=\dfrac{6}{7}(mol)$
$V_{Cl_2}=\dfrac{6}{7}.22,4=19,2l$
$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O$
$\Rightarrow n_{Cl_2}=\dfrac{5}{16}.4=1,25(mol)$
$V_{Cl_2}=1,25.22,4=28l$
b,
$Fe+2HCl\to FeCl_2+H_2$
(thiếu đề, không tính m được)