Respuesta :
First calculate the number of moles of bromine and benzene.
Number of moles = [tex]\frac{given mass in g}{molar mass}[/tex]
Mass of benzene = [tex]32.0 g[/tex] (given)
Molar mass of benzene= [tex]78.11 g/mol[/tex]
Substitute the above values in formula, we get
Number of moles of benzene= [tex]\frac{32.0 g}{78.11 g/mol}[/tex]
= [tex]0.409 mol[/tex]
Mass of bromine = 69.3 g
Molar mass of bromine = [tex]2\times 79.9 g/mol[/tex] = [tex]159.8 g/mol[/tex]
Substitute the above values in formula, we get
Number of moles of benzene= [tex]\frac{69.3 g}{159.8 g/mol}[/tex]
= [tex]0.43 g/mol[/tex]
Now, ratio of benzene and bromine comes out be 1:1 respectively and limiting reagent is benzene because present less as compared to bromine.
Thus, number of moles of bromobenzene = [tex]0.409 mol[/tex]
Theoretical yield = [tex]number of moles \times molar mass[/tex]
= [tex]0.409 mol \times 157.02 g/mol[/tex] (molar mass of bromonenzene = 157.02 g/mol)
= [tex]64.22118 g[/tex]
Hence, theoretical yield of bromobenzene is [tex]64.22118 g[/tex]
