Answer:
71.7 g of C₆H₅Br are produced at the theoretical yield.
Explanation:
The reaction is:
C₆H₆ + B₂ → C₆H₅Br + HBr
This is an easy excersie, stoichiometry is 1:1
We determine moles of reactants:
42.1 g / 78g/mol = 0.540 moles of benzene
73 g / 159.80 g/mol = 0.457 moles of hydrogen bromide
Certainly, the HBr is the limiting reactant, because I need 0.540 moles of HBr for 0.540 moles of benzene and I only have, 0.457 moles.
0.457 moles of HBr will produce 0.457 of bromobenzene, at the 100 % yield reaction (theoretical yield)
We convert the moles to mass: 0.457 g . 157 g/mol = 71.7 g
The theoretical yield of the reaction has been 0.457 mol.
The balanced chemical equation for the reaction can be given by:
[tex]\rm C_6H_6\;+\;Br_2\;\rightarrow\;C_6H_5Br\;+\;HBr[/tex]
Accordingly, 1 mole of benzene will give 1 mole of Bromobenzene and 1 mole of Hydrobromide.
The moles can be calculated as;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles in 42.1 grams Benzene:
Moles of Benzene = [tex]\rm \dfrac{42.1}{78}[/tex]
Moles of Benzene = 0.540 mol
Moles of Bromine = [tex]\rm \dfrac{73}{159.80}[/tex]
Moles of Bromine = 0.457 mol.
The limiting reactant in the reaction has been the one that has been present in the lesser quantity as the other reactant to continue to the product formation.
The 1 mole of Benzene requires 1 mole of bromine. So, 0.54 mol of benzene requires. 0.54 mol of bromine. Since the available bromine has been 0.457 mol, Bromine has been the limiting reactant.
The theoretical yield of the reaction will be based on the concentration of Bromine.
Thus, 1 mol of Bromine forms 1 mol of Bromobenzene.
0.457 mol of Bromine will form 0.457 mol of bromobenzene.
The theoretical yield of the reaction has been 0.457 mol.
For more information about theoretical yield, refer to the link:
https://brainly.com/question/12704041
