Question

How many moles of hcl are required to neutralize aqueous solutions of these bases:

a.) 0.03 mol koh
b.) 2 mol nh3
c.) 0.1 mol ca(oh)2?

Answer 1

a) Chemical reaction: KOH + HCl → KCl + H₂O.
n(HCl) : n(KOH) = 1 : 1.
n(HCl) = 0.03 mol.
b) Chemical reaction: HCl + NH₃ → NH₄Cl.
n(HCl) : n(NH₃) = 1 : 1.
n(HCl) = 2 mol.
c) Chemical reaction: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O.
n(HCl) : n(Ca(OH)₂) = 2 : 1.
n(HCl) = 0.2 mol.

Answer 2

The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.

b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.

c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.