Answer: 131 g of bromine is required.
Explanation:
The balanced equation will be :
[tex]2AlCl_3+3Br_2\rightarrow 2AlBr_3+3Cl_2[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
moles of [tex]AlCl_3[/tex]
[tex]\text{Number of moles}=\frac{72.4g}{133g/mol}=0.544moles[/tex]
According to stoichiometry :
2 moles of [tex]AlCl_3[/tex] require = 3 moles of [tex]Br_2[/tex]
Thus 0.544 moles of [tex]AlCl_3[/tex] require=[tex]\frac{3}{2}\times 0.544=0.816moles[/tex] of [tex]Br_2[/tex]
Mass of [tex]Br_2=moles\times {\text {Molar mass}}=0.816moles\times 160g/mol=131g[/tex]
Thus 131 g of bromine is required.
