Dr. S. N. Too ran the above reaction using 10.00 g of 1,5-pentanediol and 7.00 g

Question

of HCl. He obtained 9.23 g of 1,5-dichloropentane. What was his per cent yield?
HO(CH2)5OH + 2 HCl --------> Cl(CH2)5Cl + 2 H2O

Answer 1

This is a limiting reagent problem.
You have the equation.
Convert 10.00 g of 1,5 pent--- to mols. mols = grams/molar mass

Convert 7.00g HCl to mols the same way.

Using the coefficients in the balanced equation convert mols of each reactant to mols of the product. It is likely you will get two numbers; of course that can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.
Then convert the smaller number into grams. g = mols x molar mass. This is the theoretical yield (TY). The actual yield (AY) is 9.23 g
%yield = (AY/TY)*100 = ?