Question

Consider the neutralization between the following two solutions:
100.0 mL 0.9285 M hydrochloric acid, HCl
150.0 mL 1.8269 M sodium hydroxide, NaOH

How many moles of water are produced by this neutralization?

Answer 1

HCl + NaOH ==> NaCl + H2O
100.0 mL 0.9285 M hydrochloric acid, HCl
150.0 mL 1.8269 M sodium hydroxide, NaOH

mols HCl initially = mL/1000 x M = 0.09285
mols NaOH initially = 0.2740
mols H2O produced IF we use HCl as the limiting reagent (LR) = 0.09285
mols H2O produced IF we used NaOH as the LR = 0.2740
In LR problem we always produce the LEAST amount possible and have an excess of the other reagent; i.e., we will have an excess of NaOH.