$n_{MnO_2}=\dfrac{69,6}{87}=0,8(Mol)$

Đổi $500ml=0,5l$

$\Rightarrow n_{NaOH}=0,5.4=2(mol)$

$PTHH:MnO_2+4HCl\to MnCl_2+Cl_2\uparrow+2H_2O(1)$

$Cl_2+2NaOH\to NaCl+NaClO+H_2O(2)$

Theo PT(1): $n_{Cl_2}=n_{MnO_2}=0,8(mol)$

Xét tỉ lệ PT(2): $n_{Cl_2}<\dfrac{n_{NaOH}}{2}\Rightarrow NaOH$ dư sau p/ứ (2)

Theo PT(2): $n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,8(mol);n_{NaOH}=2n_{Cl_2}=1,6(mol)$

$\Rightarrow n_{NaOH(dư)}=2-1,6=0,4(mol)$

$V_{\text{dd sau}}=V_{dd_{NaOH}}=0,5(l)$

$\Rightarrow \begin{cases} C_{M_{NaOH}}=\dfrac{0,4}{0,5}=0,8M\\ C_{M_{NaCl}}=\dfrac{0,8}{0,5}=1,6M\\ C_{M_{NaClO}}=\dfrac{0,8}{0,5}=1,6M \end{cases}$

Đáp án: Bla bla

MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O
Ta có : nMnO2=69,6/(55+16.2)=0,8 mol
Theo ptpu: nCl2=nMnO2=0,8 mol
nNaOH=0,5.4=2 mol
Cho Cl2 vào dung dịch NaOH
2NaOH + Cl2 -> NaCl + NaClO + H2O
Vì nNaOH > 2nCl2 nên NaOH dư
-> nNaOH phản ứng=2nCl2=0,8.2=1,6 mol
-> nNaOH dư=2-1,6=0,4 mol
nNaCl=nNaClO=nCl2=0,8 mol
->CM NaOH dư=0,4/0,5=0,8M
CM NaCl= CM NaClO=0,8/0,5=1,6M

Bạn tham khảo nha ^^