See your other post for part a.
b. Convert 2.50 what, grams? to mols of the dibromo cmpd, convert to mols C6H6 used, convert to grams C6H6 used. This is the amount not converted to the monobromo cmpd.
c. Amount used for conversion to the monobromo cmpd = initial amount of 15.0-amount used in part b. Now go through the calculations as I did in the other post for theoretical yield to find the actual yield (AY). The other post is the theoretical yield (TY)
d. % yield = (AY/TY)*100 = ?
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?
1 answer
1 answer