What is the percentage yield when 2.0g of ethene, c2h4, is formed from 5g of ethanol, c2h5oh

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 %

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%