Đáp án:
a) C% HCl=5,85%; C% MnCl2=20,18%
b)
CM NaCl= CM NaClO=0,6M
CM NaOH dư=0,8M
Giải thích các bước giải:
a) MnO2 + 4HCl ->MnCl2 + Cl2 +2H2o
Ta có: nMnO2=2,61/(55+16.2)=0,03 mol
mHCl=18,25.30%=5,475 gam -> nHCl=0,15 mol
Vì nHCl >4nMnO2 -> HCl dư
-> nHCl dư=0,15-0,03.4=0,03 mol; nCl2=nMnCl2=nMnO2=0,03 mol
BTKL: m dung dịch A=2,61+18,25-mCl2=2,61+18,25-0,03.71=18,73 gam
mHCl dư=1,095 gam; mMnCl2=3,78 gam
-> C% HCl=5,85%; C% MnCl2=20,18%
b) Cl2 + 2NaOH -> NaCl + NaClO + H2O
Ta có: nNaOH=0,05.2=0,1 mol >2nCl2
-> NaOH dư -> nNaOH dư=0,1-0,03.2=0,04 mol.
nNaClO=nNaCl=0,03 mol
-> CM NaCl= CM NaClO=0,03/0,05=0,6M
CM NaOH dư=0,04/0,05=0,8M