Question

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of water formed from the reaction of 14.9 g of hydrochloric acid and 26.9 g of sodium hydroxide?

Answer 1

First, we write the chemical equation described:

HCl (aq) + NaOH (s) -> NaCl (aq) + H₂O (l)

Then, we are told that we start with 14.9g of HCl and 26.9 of sodium hydroxide. We turn them to moles (with their molar masses) so that we may determine the limiting reactant:

14.9 g of HCl * (1 mol / 36.46 grams) = 0.4087 moles of HCl

26.9 g of NaOH * (1 mol / 39.99 grams) = 0.6727 moles of NaOH

As the stoichiometric relation is 1 on 1, the limiting reactant is hydrochloric acid. Also, as the relation with water is 1 on 1, we will produce 0.4087 moles of water, which we can turn to the theoretical yield with its molar mass:

0.4087 moles of H₂O * (18.01 g / 1 mol) = 7.36 grams of Water