If you have 10 mL of HCl and 50mL of NaOH, how many moles

ofwater are produced by the acid-base reaction?

0.3233 mol

Explanation:

Let's suppose that the acid (HCl) is concentrated (almost pure), thus, its density is 1.18 g/mL. The density of NaOH is 2.13 g/mL. The molar masses are: HCl = 36.5 g/mol, NaOH = 40 g/mol.

When an acid and a base react, they are neutralized, and salt and water are formed. The reaction of these two substances is:

HCl + NaOH → NaCl + H₂O

Thus, by the stoichiometry, 1 mol of HCl must react with 1 mol of NaOH to form 1 mol of water (H₂O). Let them found out which of the reactants is limiting. The mass is the volume multiplied by the density:

mHCl = 10 mL * 1.18 g/mL = 11.8 g

mNaOH = 50 mL * 2.13 g/mL = 106.5 g

The number of moles is the mass divided by the molar mass:

nHCl = 11.8/36.5 = 0.3233 mol

nNaOH = 106.5/40 = 2.6625 mol

So, HCl is the limiting reactant, thus it will react completely, and 0.3233 mol of water will be formed.