a)
$Cl_2 + 2NaOH \to NaCl + NaClO + H_2O$
b)
$n_{Cl_2} = \dfrac{1,12}{22,4} = 0,05(mol)$
Theo PTHH : $n_{NaOH} = 2n_{Cl_2} = 0,1(mol)$
$V = \dfrac{0,1}{0,1} = 1(lít)$
c) $n_{NaCl} = n_{NaClO} = n_{Cl_2} = 0,05(mol)$
$C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,05}{1} = 0,05(M)$