percentage yield

good question.

now, assume the percentage yield is 100%, i.e. the reaction is complete.
then, how many mole of bromoethane CH3CH2Br can you get?
0.071 mole (for ethene)? 0.025 mole (for HBr)?

you see the point.
even when the reaction is complete, we don't consider the ethene, which is in excess.
the MAX. no. of mole of product formed is determined by the reactant with LEAST amount, so-called "limiting reagent".
in this case, ethene is the limiting reagent, while hydrogen bromide is in excess.

in such calculation of percentage yield, we only consider the limiting reagent.


therefore, it's important to:
1. write down the equation of reaction.
2. calculate and write down the no. of mole of ALL reactants.

3. from the equation (the coefficients in the eq.) determine which one is limited. [there is possibility that no one is in excess, i.e. reactants are just enough for reaction. in this case, the reactants are called in stoichiometry; it's a stoichiometric reaction]

4. from equation and no. of mole of limiting reagent, calculate theoretical (maximum) no. of mole and/or mass of products formed.

5. given the actual no. of mole and/or mass of product, you now get the answer.