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:

C_{6}H_{6} + Br_{2} -------> C_{6}H_{5}Br + 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**:
S_{7}P_{6}A_{66} (grouping/interaction - chemical).

S_{7}P_{1}A_{15} (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

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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