Determining Percent Yield Of BromoBenzene From Benzene And Bromine Reaction
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Determining Percent Yield Of BromoBenzene From Benzene And Bromine Reaction Bromobenzene is obtained when benzene and bromine react as follows:

C6H6 + Br2 -------> C6H5Br + HBr

(a) What is the theoretical yield of bromobenzene in this reaction when 30 g of benzene reacts with 65 g of bromine?
(b) What is the percent yield of bromobenzene if the actual yield of bromobenzene is 56.7 g?

The strings: S7P6A66 (grouping/interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating, mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type grouping/interaction (chemical). (a) Molar mass of benzene = 12x6 + 1x6 = 78 g
78 g is the weight of 1 mole of benzene
So, 30 g is the weight of 30/78 = o.385 mol of benzene

Molar mass of bromine = 80x2 = 160 g
160 g is the weight of 1 mole of bromine
So, 65 g is the weight of 65/160 = 0.406 moles of bromine.

So, benzene is the limiting reactant
From balanced equation 1 mole of benzene formed 1 mole of bromobenzene
So, 0.385 mol of benzene will form 0.385 mol bromobenzene

Molar mass of bromobenzene = 12x6 + 1x5 + 80 = 157 g
So, 0.385 mol of bromobenzene weighs 157(0.385) = 60.5 g
So, theoretical yield of bromobenzene is 60.5 g

(b) percent yield is = 100(actual yield/theoretical yield)
Theoretical yield of bromobenzene is 60.5 g
Actual yield of bromobenzene is 56.7 g
So, percent yield = 100(56.7/60.5) = 93.72%.

Math

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