a,
$MnO_2+ 4HCl \rightarrow MnCl_2+ Cl_2+ 2H_2O$
$Cl_2+ 2NaOH \rightarrow NaCl+ NaClO+ H_2O$
b,
$n_{Cl_2}= n_{MnO_2}= \frac{69,6}{87}= 0,8 mol$
$n_{NaOH}= 0,5.4= 2 mol$
=> Sau pứ, dd gồm:
$n_{NaCl}= n_{NaClO}= 0,8 mol$
=> $C_{M_{NaCl}}= C_{M_{NaClO}}= \frac{0,8}{0,5}= 1,6M$
$n_{NaOH dư}= 0,4 mol$
=> $C_{M_{NaOH}}= \frac{0,4}{0,5}= 0,8M$