Respuesta :
Answer:
65.8%
Explanation:
First we'll begin by writing a balanced equation for for the production of C2H4 from C2H5OH.
When ethanol (C2H5OH) undergoes dehydration, it will produce ethene (C2H4) according to the equation given below:
C2H5OH —> C2H4 + H2O
Now, from the balanced equation above, we can obtain the theoretical yield of ethene as follow:
Molar Mass of C2H5OH = (12x2) + (5x1) + 16 + 1 = 24 + 5 + 16 + 1 = 46g/mol
Molar Mass of C2H4 = (12x2) + (4x1) = 24 + 4 = 28g/mol
From the balanced equation above,
46g of C2H5OH produced 28g of C2H4.
Therefore, 5g of C2H5OH will produce = (5x28)/46 = 3.04g
Therefore, the theoretical yield of C2H4 is 3.04g
From the question, the actual yield of C2H4 is 2g
With the above information, we can easily find the percentage yield as follow
%yield = Actual yield /Theoretical yield x100
%yield = 2/3.04 x100
% yield = 65.8%
Therefore, the percentage yield of C2H4 is 65.8%
Answer:
The percentage yield of the reaction is 65.8 %
Explanation:
Step 1: Data given
Mass of ethene (C2H4) = 2.0 grams
Molar mass ethene = 28.05 g/mol
Mass of ethanol formed (C2H5OH) = 5.0 grams
Molar mass of ethanol = 46.07 g/mol
Step 2: The balanced equation
C2H5OH (aq) → C2H4 (g) + H2O (l)
Step 3: Calculate moles ethene
Moles ethanol = mass ethanol / molar mass ethanol
Moles ethanol = 5.0 grams / 46.07 g/mol
Moles ethanol = 0.1085 moles
Step 4: Calculate moles ethene
For 1 mol ethanol we'll have 1 mol ethene and 1 mol H2O
For 0.1085 moles we'll have 0.1085 moles ethene
Step 5: Calculate mass ethene
Mass ethene = 0.1085 moles * 28.05 g/mol
Mass ethene = 3.04 grams
Step 6: Calculate percent yield of the reaction
% yield = (actual mass / theoretical mass) * 100 %
% yield = (2.0 grams / 3.04 grams ) * 100 %
% yield = 65.8 %
The percentage yield of the reaction is 65.8 %
