a,
$MnO_2+4HCl\to MnCl_2+Cl_2+2H_2O$
$Cl_2+2NaOH\to NaCl+NaClO+H_2O$
b,
$n_{MnO_2}=\dfrac{69,6}{87}=0,8(mol)$
$\to n_{Cl_2}=n_{MnO_2}=0,8(mol)$
$V_{dd NaOH}=500ml=0,5l$
$\to n_{NaOH}=0,5.4=2(mol)$
$\to NaOH$ dư, $Cl_2$ hết
$n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,8(mol)$
$\to C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,8}{0,5}=1,6M$
$n_{NaOH\text{dư}}=2-0,8.2=0,4(mol)$
$\to C_{M_{NaOH}}=\dfrac{0,4}{0,5}=0,8M$