\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\)
\(n_{NaBr}=a\left(mol\right)\)
\(\Rightarrow n_{NaCl}=a\left(mol\right)\)
\(m_{hh\left(X\right)}=m\left(g\right)\)
\(\Rightarrow103a-58,5a=10\%m\)
\(\Rightarrow a=\frac{m}{445}\left(mol\right)\)
\(\%m_{NaBr}=\frac{\frac{m}{445}.103}{m}.100\%=23,15\%\)