Question

Ethyl alcohol is prepared industrially by the reaction of ethylene, C2H4, with water. What is the percent yield of the reaction if 4.6 g of ethylene gives 4.7 g of ethyl alcohol?
C2H4(g) + H2O(l) = C2H6O(l)

Answer 1

Answer:

% yield = 62.21 %

Explanation:

• C2H4(g) + H2O(l) → C2H6O(l)

∴ mass C2H4(g) = 4.6 g

∴ mass C2H6O(l) = 4.7 g

• % yield = ((real yield)/(theoretical yield))×100

theretical yield:

∴ molar mass C2H4(g) = 28.05 g/mol

⇒ mol C2H4(g) = (4.6 g)*(mol/28.05 g) = 0.164 mol

⇒ mol C2H6O(l) = (0.164 mol C2H4)*(mol C2H6O(l)/molC2H4(g))

⇒ mol C2H6O(l) = 0.164 mol

∴ molar mass C2H6O(l) = 46.07 g/mol

⇒ mass C2H6O(l) = (0.164 mol)*(46.07 g/mol) = 7.55 g

⇒ theoretical yield = 7.55 g

⇒ % yield = (4.7 g)/(7.55 g))*100

⇒ % yield = 62.21 %

Answer 2

The percentage yield of ethyl alcohol has been 62.40%.

Percentage yield can be given as the yield of the product with respect to the theoretical yield.

Percent yield can be given as:

Percentage yield = $\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100$

The theoretical yield of ethyl alcohol has been:

Moles ca be given as:

Moles = $\rm \dfrac{mass}{molecular\;mass}$

Moles of 4.6g ethylene has been:

Moles of ethylene = $\rm \dfrac{4.6}{28.05}$

Moles of ethylene = 0.16 moles.

From the balanced chemical equation,

1 mole ethylene = 1 mole ethyl alcohol

0.16 mol ethylene = 0.16 mol ethyl alcohol

The theoretical yield of ethyl alcohol = 0.16 mol.

The mass of ethyl alcohol = moles × molecular mass

The mass of ethyl alcohol = 0.16 × 46.07

The mass of ethyl alcohol = 7.3712 grams

The theoretical yield of ethyl alcohol = 7.3712 g

The actual yield of ethyl alcohol = 4.7 g

The percentage yield of reaction can be given by:

Percentage yield = $\rm \dfrac{4.7}{7.3712}\;\times\;100$

Percentage yield = 62.40%

The percentage yield of ethyl alcohol has been 62.40%.

For more information about the percentage yield, refer to the link:

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