Đáp án:
mNaOH dư=1,5 gam -> C% NaOH=4,24%
mNaCl=8,775 gam -> C% NaCl=24,845%
mNaClO3=2,6625 gam -> C% NaClO3=7,54%
Giải thích các bước giải:
2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5CL2 +8H2O
Ta có: nKMnO4=4,7/158=0,03 mol -> nCl2=5/2nKMnO4=0,075 mol
Ta có: mNaOH=30.25%=7,5 gam -> nNaOH=7,5/40=0,1875 mol
Phản ứng:
6NaOH + 3Cl2 ->5 NaCl + NaClO3 + 3H2O
-> nNaOH phản ứng=2nCl2=0,15 mol -> nNaOH dư=0,0375 mol
nNaCl=5/3nCl2=0,125 mol -> nNaClO3=1/3nCl2=0,025 mol
BTKL: m dung dịch sau phản ứng=30+mCl2=30+71.0,075=35,325 gam
mNaOH dư=1,5 gam -> C% NaOH=4,24%
mNaCl=8,775 gam -> C% NaCl=24,845%
mNaClO3=2,6625 gam -> C% NaClO3=7,54%