What mass of 2,4,6-tribromophenol (M = 331 g/mol) forms when 9.4 g of phenol (M = 94 g/mol) reacts completely with bromine water?
Correct answer D. 33.1 g
Explanation
Moles of phenol = 9.4/94 = 0.10 mol. Each mole gives one mole of C₆H₂Br₃OH, so mass = 0.10 × 331 = 33.1 g. The value 17.3 g wrongly assumes only monobromination.
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