\(n_{MnO_2}=\dfrac{1,74}{87}=0,02\left(mol\right)\)
nHCl = 2.0,2 = 0,4 (mol)
PTHH: MnO2 + 4HCl -to-> MnCl2 + Cl2 + 2H2O
0,02--->0,08----->0,02
=> \(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,4-0,08}{0,2}=1,6M\\C_{M\left(MnCl_2\right)}=\dfrac{0,02}{0,2}=0,1M\end{matrix}\right.\)