Chemical Reaction Yield Calculation: How Much Ethyl Alcohol is Needed to Produce Diethyl Ether?

How much ethyl alcohol is needed to produce 110.1 grams of diethyl ether with a 76.4% yield?

Given data: Diethyl ether (C4H10O) is prepared commercially by the treatment of ethyl alcohol (C2H6O) with an acid.

Answer:

To produce 110.1 grams of diethyl ether with a 76.4% yield, 84.174 grams of ethyl alcohol would be needed.

Chemical reactions can be very interesting to study, especially when it comes to calculating the amount of reactants needed to produce a certain product. In this case, we are looking at the production of diethyl ether from ethyl alcohol.

First, let's understand how percent yield is calculated in a chemical reaction. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. It gives us an idea of how efficient a reaction is in producing the desired product.

In our scenario, we are given that the percent yield of the reaction is 76.4%. This means that out of the theoretical yield of diethyl ether, we can expect to obtain 76.4% of it in actuality.

Using the molar ratios from the balanced chemical equation of the reaction between ethyl alcohol and diethyl ether, we can calculate the amount of ethyl alcohol needed to produce 110.1 grams of diethyl ether.

Given that 1 mole of ethyl alcohol produces 1 mole of diethyl ether, we can equate the theoretical yield of diethyl ether (110.1 grams) to the actual yield using the percent yield formula:

Actual Yield = Percent Yield × Theoretical Yield

Actual Yield = 76.4% × 110.1 g = 84.174 g

Therefore, to produce 110.1 grams of diethyl ether with a percent yield of 76.4%, we would need 84.174 grams of ethyl alcohol.

Understanding these calculations can help chemists optimize reactions and minimize waste in chemical processes. It showcases the importance of theoretical and actual yields in the realm of chemical reactions.