Ta có : \({m_{B{r_2}}} = 3,12.3 = 9,36\,\left( {kg} \right)\)
\(\Rightarrow {n_{B{r_2}}} = {{9,36.1000} \over {160}} = 58,5\,\,\left( {mol} \right)\)
\(\eqalign{ & 2NaBr + C{l_2}\,\, \to \,\,2NaCl + B{r_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \cr & 117\,\,\, \leftarrow \,\,\,58,5\,\, \leftarrow \,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;58,5 \cr} \)
Từ \(\left( 1 \right) \Rightarrow {V_{NaBr}} = {{117.103} \over {40}} = 301,3\,\,\left( {lit} \right);\)
\({V_{C{l_2}}} = 58,5.22,4 = 1310,4\,\,\left( {lit} \right)\)
