\(\begin{array}{l}
{n_{C{l_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\left( {mol} \right)\\
C{l_2} + 2KBr \to 2KCl + B{r_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
Theo\,\,\,(1)\,:\,\,\,\,\,\,{n_{KBr}} = 2.{n_{C{l_2}}} = 2.0,2 = 0,4\left( {mol} \right)\\
{m_{KBr}} = 119.0,4 = 47,6\left( g \right)\\
{m_{ddKBr}} = 88,81.1,34 = 119\left( g \right)\\
C\% = \dfrac{{47,6}}{{119}}.100\% = 40\%
\end{array}\)
