Question

4. A reaction is carried out for which the stoichiometry is A + 2B + 3C → D + 2E + F. In this reaction, 0.5 mole of A, 0.75 mole of B, and 1.0 mole of C are allowed to react. The yield of E is 0.2 mole. What is the percentage yield?

Answer 1

You have to start by identifying the "limiting reactant", which is other words it's the substance you have least of taking into account the stoichiometric coefficients.

A + 2B + 3C → D + 2E + F

For example, there are 0.75 moles of B but when you take into account the number of molecules interacting in the reaction, you have 0.75/2 = 0.375 moles

notice that the 2 came from the balanced equation above.

You do this for all the reactants and determine which has the least amount of moles.

Now, to find the percentage yield, you need the theoretical yield, which is the greatest amount you can have from the reaction (which never occurs due to real life complications). Then apply:

percent yield = (actual yield/theoretical yield)*100% (equation 1)

To find the theoretical yield you use typical stoichiometric calculations, where you use a ratio comparing the 2 substances of interest, the limiting reactant and the product (E).

A + 2B + 3C → D + 2E + F

so: moles of limiting reactant/coefficient=moles of E/2

(you need to determine which is the limiting reactant first)

So, find how many moles of E you can produce. Now plug (into equation 1) the actual yield (given in the question) and the theoretical yield to determine the percent yield.