IMAT 2015 Q42 [Percentage Yield]

In the reaction

C_3H_7Br + KOH → C_3H_7OH + KBr

24.6 g of 1 ­bromopropane reacts with excess potassium hydroxide to produce 8.00 g of propan­-1-­ol.

M_r : C3H7Br = 123
A_r : H = 1.0; C = 12.0; O = 16.0

What is the percentage yield of this reaction?

A. 57.1%
B. 93.0%
C. 32.5%
D. 33.3%
E. 66.7%