Question

Ethanol (CH3CH2OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuric acid, ethanol forms diethyl ether and water: 2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(g) In a side reaction, some ethanol forms ethylene and water: CH3CH2OH(l) → CH2═CH2(g) + H2O(g) a) If 50.0 g of ethanol yields 35.9 g of diethyl ether, what is the percent yield of diethyl ether? b) During the process, 45.0% of the ethanol that did not produce diethyl ether reacts by the side reaction. What mass of ethylene is produced?

Answer 1

a) 88.48%

b) 0.05625 mol

Step-by-step explanation:

2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1

CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2

a) CH₃CH₂OH = 46.0684 g/mol

CH₃CH₂OCH₂CH₃ = 74.12 g/mol

1 mol CH₃CH₂OH ______ 46.0684 g

x ______ 50.0 g

x = 1.085 mol CH₃CH₂OH

1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g

y ______ 35.9 g

y = 0.48 mol CH₃CH₂OCH₂CH₃

100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃

w _____ 0.48 mol CH₃CH₂OCH₂CH₃

w = 88.48%

b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.