% yield = 74.35
<h3>Further explanation</h3>
Given
12 dm³ ethene = 12 L
18.4 g ethanol(actual)
Required
The percentage yield
Solution
Reaction
C₂H₄(g) + H₂O(g) ⇒ C₂H₅OH(g)
Assume at STP, 1 mol gas = 22.4 L
mol ethene :
= 12 L : 22.4 L
= 0.538
From equation, mol ratio C₂H₅OH : C₂H₄ = 1 : 1, so mol C₂H₅OH = 0.538
Mass Ethanol (MW=46 g/mol) ⇒ theoretical
= 0.538 x 46
= 24.748 g
% yield = (actual/theoretical) x 100%
% yield = (18.4/24.748) x 100%
% yield = 74.35
