$n_{MnO_2}=\dfrac{12,6}{87}\approx 0,14(mol)\\ n_{HCl}=2.0,25=0,5(mol)\\ a,PTHH:MnO_2+4HCl\xrightarrow{t^o} MnCl_2+Cl_2+2H_2O\\ b,\text{Lập tỉ lệ: }\dfrac{0,14}{1}>\dfrac{0,5}{4}\Rightarrow MnO_2\text{ dư}\\ \Rightarrow n_{Cl_2}=\dfrac{1}{4}n_{HCl}=0,125(mol)\\ \Rightarrow V_{Cl_2}=0,125.22,4=2,8(l)\\ c,n_{MnCl_2}=n_{Cl_2}=0,125(mol)\\ \Rightarrow m_{MnCl_2}=0,125.126=15,75(g)$

`#Ly`

$a)$

PTHH: $MnO_2$ $+$ $4HCl$ `->` $MnCl_2$ $+$ $Cl_2↑$ $+$ $2H_2O$

$b)$

`n_(MnO_2` `=` `m/M` `=` `(12,6)/(87)` `=` `(21)/(145)` $(mol)$

Đổi $250$ $ml$ $=$ $0,25$ $l$

`n_(HCl` `=` `C_M.V=0,25.2=0,5` $(mol)$

Ta có: `(21)/(145)` $>$ $0,5$

`=>` $MnO_2$ dư, $HCl$ pư hết nên tính theo `n_(HCl`

Theo PT, `n_(Cl_2` `=` `(1)/(4)n_(HCl `=` $0,125$ $(mol)$

`V_(Cl_2` `=` `n.22,4=0,125.22,4=2,8` $(l)$

$c)$

Theo PT, `n_(MnCl_2` `=` `(1)/(4)n_(HCl `=` $0,125$ $(mol)$

`m_(MnCl_2` `=` $n.M=0,125.126=15,75$ $(g)$