\(a,n_{MnO_2}=\dfrac{34,8}{87}=0,4\left(mol\right)\)
PTHH: MnO2 + 4HCl ---> MnCl2 + 2H2O + Cl2
0,4------------------>0,4--------------->0,4
=> VCl2 = 0,4.22,4 = 8,96 (l)
b) \(n_{NaOH}=0,5.2=1\left(mol\right)\)
PTHH: 2NaOH + Cl2 ---> NaCl + NaClO + H2O
LTL: \(\dfrac{1}{2}>0,4\) => NaOH dư
PTHH: 2NaOH + Cl2 ---> NaCl + NaClO + H2O
0,8<----0,4------>0,4------>0,4
=> \(\left\{{}\begin{matrix}C_{M\left(NaOH\right)}=\dfrac{1-0,8}{0,5}=0,4M\\C_{M\left(NaCl\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(NaClO\right)}=\dfrac{0,4}{0,5}=0,8M\end{matrix}\right.\)