\(2NaBr + Cl_2 \to 2NaCl + Br_2\\ 2NaI + Cl_2 \to 2NaCl + I_2\\ n_{Cl_2} =\dfrac{1}{2}n_{NaCl} = \dfrac{1}{2}.\dfrac{23,4}{58,5} = 0,2(mol)\\ \Rightarrow V_{Cl_2} = 0,2.22,4 = 4,48(lít)\)
\(n_{NaCl}=\dfrac{23.4}{58.5}=0.4\left(mol\right)\)
\(BTNTCl:\)
\(n_{Cl_2}=\dfrac{n_{NaCl}}{2}=\dfrac{0.4}{2}=0.2\left(mol\right)\)
\(V_{Cl_2}=0.2\cdot22.4=4.48\left(l\right)\)