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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