In the laboratory, from 129 grams of chloroethane, 85 grams of ethanol was obtained.

In the laboratory, from 129 grams of chloroethane, 85 grams of ethanol was obtained. How much is this% of the theoretically possible?

Given:
m (CH3-CH2-Cl) = 129 g
m practical (CH3-CH2-OH) = 85 g
To find:
Yield (CH3-CH2-OH)
Decision:
CH3-CH2-Cl + KOH (aq) = KCl + CH3-CH2-OH
n (CH3-CH2-Cl) = m / M = 129 g / 64.5 g / mol = 2 mol
n (CH3-CH2-Cl): n (CH3-CH2-OH) = 1: 1
n (CH3-CH2-OH) = 2 mol
m theor (CH3-CH2-OH) = n * M = 2 mol * 46 g / mol = 92 g
Yield (CH3-CH2-OH) = m practical / m theory * 100% = 85 g / 92 g * 100% = 92.4%
Answer: 92.4%