2.74g of 1-bromobutane is reacted with excess aqueous sodium hydroxide to produce 1.11g of butan-1-ol according to the equation below.
C4H9Br + NaHO → C4 H9OH + NaBr
What is the percentage yield of butan-1-ol?

Answer:

2) D

Mr of 1-bromobutane = (12×4) + (1×9) + 80 = 137
Moles of 1-bromobutane = mass/Mr = 2.74/137 = 0.02 mol

Molar ratio of 1-bromobutane and butan-1-ol = 1:1
Moles of butan-1-ol = 0.02 mol
Mr of butanol = (12×4) + (1×9) + (16) + (1) = 74
Mass of butanol = moles x Mr = 0.02 x 74 = 1.48 g
Mass of butan-1-ol obtained = 1.11g

Percentage yield = (1.11/1.48) x 100 = 75%
Hence, the answer is D