Question

Ethyl alcohol ${{C}_{2}}{{H}_{5}}OH$ is prepared industrially by reaction of ethylene ${{C}_{2}}{{H}_{4}}$ with water. What is the percent yield of the reaction if 9.6g of ethylene gives 4.7 g of ethyl alcohol. (Round-off the answer to the nearest integer.)
\[{{C}_{2}}{{H}_{4}}+{{H}_{2}}O\to {{C}_{2}}{{H}_{5}}OH\]

Answer 1

Hint: Percentage yield is ratio of actual yield to theoretical yield. The theoretical yield is the maximum amount of product that is obtained from a reaction. Actual yield is the amount of product actually produced in reaction. Molecular weight of ethyl alcohol is 46 and molecular weight of ethylene is 28.

Complete answer:
-1 mole of ethylene reacts with 1 mole of water to produce 1 mole of ethyl alcohol
- molar mass of ethylene is 28, so mole of ethyl alcohol can be calculated.
Moles of Ethylene= $\dfrac{Given\text{ mass}}{molar\text{ mass}}=\dfrac{9.6}{28}$=0.3428moles
As 1 mole of Ethylene produces 1 mole of Ethyl alcohol.
So, 0.3428 moles of Ethylene will produce 0.3428 moles of Ethyl Alcohol.
Molar mass of Ethyl Alcohol=46
Moles of Ethyl alcohol=\[\dfrac{\text{Given Mass}}{\text{Molar mass}}=\dfrac{4.7}{46}\]=0.10217 moles
So, Actual yield is 0.10217 moles and theoretical yield is 0.3428 moles.
So, Percent yield can be calculated as
=$\dfrac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100=\dfrac{0.10217}{0.3428}\times 100$=29.80=30%
So Percent yield of reaction is 30%.

Note:
Yield is moles of product formed relating to moles of reactant consumed. Yield is the primary factor considered in synthesis of organic and inorganic compounds. Yields above 90% are excellent and below 40% are poor. Stoichiometry is used to determine yield of chemical reaction by determining mole ratio. It also determines limiting reagent which is completely consumed in reaction.