STOICHIOMETRY: QUANTITATIVE INFORMATION ABOUT CHEMICAL REACTIONS

Question

A 12.3 g sample of an aqueous solution of hydrochloric acid contains an unknown amount of the acid.

If 11.3 mL of 1.46 M sodium hydroxide are required to neutralize the hydrochloric acid, what is the percent by mass of hydrochloric acid in the mixture?

Answer 1

First, we'll write the balanced equation for reaction/neutralization of HCl with NaOH. Then, we'll find the moles of HCl present, convert that to mass of HCl, and finally determine the percent by mass of HCl in the mixture.

HCl + NaOH ==> NaCl + H2O

moles NaOH = 11.3 ml x 1 L/1000 ml x 1.46 mole/L = 0.016498 moles NaOH

moles HCl = 0.016498 moles NaOH x 1 mol HCl/mol NaOH = 0.016498 moles HCl

mass HCl = 0.016498 moles HCl x 36.5 g/mol = 0.602 g HCl

mass % HCl = mass HCl/total mass (x100%) = 0.602 g/12.3 (x100%) = 4.90% HCl