nKMnO4=0,03 mol
nNaOH=0,1875
PTHH:
2KMnO4+ 16HCl→2KCl+ 2MnCl2+5Cl2+8H2O
0,03__________________________0,075
3Cl2+ 6NaOH→ 5NaCl+ NaClO3+3H2O
0,075___0,15____ 0,125 ____0,025
m dd sau pư= 30+0,075.71=35,325 g
C% NaCl= \(\frac{\text{0,125.58,5}}{35,325}.100\%\)=20.7%
C%NaClO3= \(\frac{\text{0,025.106,5}}{35,325}.100\%\)=7,53%
C%NaOH dư= \(\frac{\text{0,0375.40}}{35,325}.100\%\)=4,25%