Answer:
The % yield is 62.7 %
Explanation:
<u>Step 1:</u> the balanced equation
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
<u>Step 2:</u> Data given
mass of ethanol = 52.9 grams
molar mass of ethanol = 46.07 g/mol
mass of ether = 26.7 grams
Molar mass of ether = 74.12 g/mol
<u>Step 3:</u> Calculate moles of ethanol
Number of moles = Mass / molar mass
Moles of ethanol = 52.9 grams / 46.07 g/mol
Moles of ethanol = 1.15 moles
<u>Step 4: </u>Calculate moles of ether
For 2 moles of ethanol consumed, we have 1 mole of ether produced
For 1.15 moles of ethanol consumed, we have 1.15/2 = 0.575 moles of ether produced
<u>Step 5:</u> Calculate mass of ether
Mass = moles * Molar mass
Mass ether = 0.575 moles * 74.12 g/mol = 42.62 grams = theoretical yield
<u>Step 6:</u> Calculate % yield
% yield = (26.7 / 42.62) * 100 = 62.7%
The % yield is 62.7 %
