\(n_{Cl_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right);n_{NaOH}=0,2.1,5=0,3\left(mol\right)\)
Ta có: \(\dfrac{0,125}{1}< \dfrac{0,3}{2}\) ⇒ Cl2 hết, NaOH dư
Cl2 + 2NaOH → NaCl + NaClO + H2O
Mol: 0,125 0,25 0,125 0,125
\(C_{M_{NaOHdư}}=\dfrac{0,3-0,125}{0,2}=0,875M\)
\(C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,125}{0,2}=0,625M\)