a.n MnO2= 0.15 mol
MnO2+ 4Hcl=mncl2+cl2+2h2o
0.15............................0.15(mol)
V=0.15*22.4=3.36l
V Naoh=200ml=0.2l=))nNaOh=1/120mol
2Naoh+cl2=nacl+naclo+h2o
1/120.....1/240
=)) Cl2 dư=7/48mol
Cm NaCl=NaClO=1/240/0.2=1/48M
n Koh =0.4 mol
6KOH+ 3Cl2= KClO3+5KCl+3h2o
0.3..........0.15...0.05.....0.25 =)) N KOH dư=0.1 mol
mdd KOH= 224g
mcl2=10.65g
C%KOH dư= 0.1*56/(224+10.65)*100=2.4%
C%KClO3= 0.05*122.5/(224+10.65)*100=2.6%
C% KCl= 0.25*74.5/(224+10.65)*100=7.9%