When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained

Question

When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained: C6H6+Br2C6H5Br+HBr

Part A

What is the theoretical yield of bromobenzene in this reaction when 33.0 g of benzene reacts with 71.5 g of bromine?

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Part B

If the actual yield of bromobenzene was 63.0 g , what was the percentage yield?

When benzene (C6H6) reacts with bromine (Br2), bromobenzene (C6H5Br) is obtained: C6H6+Br2C6H5Br+HBr

Part A

What is the theoretical yield of bromobenzene in this reaction when 33.0 g of benzene reacts with 71.5 g of bromine?

SubmitRequest Answer

Part B

If the actual yield of bromobenzene was 63.0 g , what was the percentage yield?

Explanation / Answer

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

78 g C6H6 reacts with 159.8 g Br2

33.0 g C6H6 reacts with 33.0 x 159.8 / 78 = 67.6 g Br2

we have 71.5 g Br2. so Br2 is exess reagent . C6H6 is limiting reagent.

78 g C6H6 gives 157 g C6H5Br

33.0 g C6H6 gives 33.0 x 157 / 78 = 66.42 g

therotical yield = 66.42 g

actual yield = 63.0 g

% yield = (63.0 / 66.42) x100

% yield = 94.8

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