In an experiment a student allowed benzene C6H6, TO REACT WITH EXCESS BROMINE BR2, in an attempt to prepare bromobenzene C6H5Br. This reaction also produced as a by-product, dibromobenzene, C6H4Br2, On the basis of the equation.
C6H6 + Br2 -> c6h5Br + H Br
a)What is the maximum amount of C6H5Br that the student could have hoped to obtain from 15.0 g of benzene? (This is the theoretical yield.)
b)In this experiment the student obtained 2.50 C6H5Br2, How much C6H6 was not converted to C6H5Br?
c)What is the student’s actual yield of C6H5Br?
d)Calculate the percentage yield for the reaction?
See your earlier post under dolly.
To solve these stoichiometry problems, we need to use the balanced chemical equation and the given information.
The balanced equation for the reaction is:
C6H6 + Br2 → C6H5Br + HBr
a) To find the maximum amount (theoretical yield) of C6H5Br that can be obtained from 15.0 g of benzene (C6H6), we need to use the molar masses and stoichiometry.
1) Calculate the molar mass of benzene (C6H6):
C6H6 = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol
2) Convert the mass of benzene to moles:
Moles of C6H6 = 15.0 g / 78.11 g/mol = 0.192 moles
3) Use stoichiometry to determine the moles of C6H5Br (bromobenzene) produced for every mole of benzene used:
According to the balanced equation, the coefficient of C6H5Br is 1, and the coefficient of C6H6 is also 1. Therefore, the moles of C6H5Br produced will be equal to the moles of the benzene used.
Moles of C6H5Br = 0.192 moles
4) Convert moles of C6H5Br to grams:
Mass of C6H5Br = Moles of C6H5Br ⨉ molar mass of C6H5Br
= 0.192 moles ⨉ (79.93 g/mol)
= 15.34 g
So, the theoretical yield of C6H5Br from 15.0 g of benzene is 15.34 g.
b) To find how much C6H6 was not converted to C6H5Br, we need to subtract the amount of C6H5Br produced (as calculated in part a) from the initial amount of benzene used.
Initial moles of C6H6 = 0.192 moles (obtained in part a)
Moles of C6H6 remaining = Initial moles of C6H6 − Moles of C6H5Br produced
= 0.192 moles − 0.192 moles
= 0 moles
Since there are 0 moles of benzene remaining, we can conclude that all of the benzene has been converted to C6H5Br.
c) The actual yield of C6H5Br is given as 2.50 g.
d) To calculate the percentage yield, we use the formula:
Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (2.50 g / 15.34 g) * 100
= 16.29 %
Therefore, the percentage yield of C6H5Br in this reaction is 16.29%.