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2 years ago

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

Chemistry

1 answer:

Blababa [14]2 years ago

5 0

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

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