Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,
30 g / 78 moles of benzene = 0.384 moles
For bromine it will be the same process,
65 g / 159.8 moles = 0.406 moles
By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1
We need to find the mass of bromobenzene,
which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol
So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g
Hence the theoretical yield will be 60.34 g
Answer B) : To calculate the actual yield we have to divide it with theoretical yield.
(56.7g / 60.343 g ) X100% = 93.96 %
Here, we can say that we got 93.96 % of actual yield.
As we know it is impossible to get 100% yield in any reaction.
The percentage yield of bromobenzene is 95%.
The equation of the reaction is;
C6H6 + Br2 → C6H5Br + HBr
Number of moles of C6H6 reacted = 30.0g/78 g/mol = 0.38 moles
Number of moles of Br2 = 65.0 g/160 g/mol = 0.41 moles
Since the reaction is 1:1, C6H6 is the limiting reactant.
The theoretical yield of bromobenzene = 0.38 moles × 157 g/mol = 59.66 g
Given that;
%yield = actual yield/theoretical yield × 100/1
Actual yield = 56.7 g
%yield = 56.7 g/59.66 g × 100/1
= 95%
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