nNaOH = 1.2,5 = 2,5 (mol)
\(n_{Cl_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2NaOH + Cl2 --> NaCl + NaClO + H2O
Xét tỉ lệ: \(\dfrac{2,5}{2}>\dfrac{0,6}{1}\) => NaOH dư, Cl2 hết
PTHH: 2NaOH + Cl2 --> NaCl + NaClO + H2O
1,2<---0,6
=> \(C_{M\left(NaOH.sau.pư\right)}=\dfrac{2,5-1,2}{2,5}=0,52M\)
=> A