Đáp án:

CMNaOH dư=1 M

CMNaCl=CMNaClO=2M

C%NaOH=2,87%

C%NaCl=8,41%

C%NaClO=10,7%

Giải thích các bước giải:

 nMnO2=87/87=1 mol

MnO2+4HCl->MnCl2+Cl2+2H2O

1 mol                            1 mol

Cl2+2NaOH->NaCl+NaClO+H2O

1mol   2mol  1mol     1 mol

nNaOH dư=0,5x5-2=0,5 mol

CMNaOH dư=0,5/0,5=1 M

CMNaCl=CMNaClO=1/0,5=2M

m dd NaOH=1,25x500=625 g

mdd spu=1x71+625=696 g

C%NaOH=0,5x40/696x100%=2,87%

C%NaCl=1x58,5/696x100%=8,41%

C%NaClO=1x74,5/696x100%=10,7%

`n_{MnO_2}={87}/{87}=1(mol)`

`MnO_2+4HCl->MnCl_2+Cl_2+2H_2O`

`n_{Cl_2}=n_{MnO_2}=1(mol)`

`n_{NaOH}=0,5.5=2,5(mol)`

`Cl_2+2NaOH->NaCl+NaClO+H_2O`

Do `2.1<2,5->NaOH` dư

`n_{NaCl}=n_{NaClO}=1(mol)`

`n_{NaOH(du)}=2,5-1.2=0,5(mol)`

`m_{dd\ spu}=500.1,25+1.71=696(g)`

`C_{M\ NaCl}=C_{M\ NaClO}={1}/{0,5}=2M`

`C_{M\ NaOH(du)}={0,5}/{0,5}=1M`

`C\%_{NaCl}={1.58,5}/{696}.100≈8,41\%`

`C\%_{NaClO}={1.74,5}/{696}.100≈10,7\%`

`C\%_{NaOH(du)}={0,5.40}/{696}.100≈2,87\%`