## How many moles of hcl are required to neutralize aqueous solutions of these bases: a.) 0.03 mol koh b.) 2 mol nh3

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?

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1. 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) u2192 KCl (aq) + H2O(u2113)
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) ) u2192 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) u2192 CaCl2(aq) + 2H2O(u2113)
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.

2. a) Chemical reaction: KOH + HClu2192 KCl + Hu2082O.
n(HCl) : n(KOH) = 1 : 1.
n(HCl) = 0.03 mol.
b) Chemical reaction: HCl + NHu2083 u2192 NHu2084Cl.
n(HCl) : n(NHu2083) = 1 : 1.
n(HCl) = 2 mol.
c) Chemical reaction: Ca(OH)u2082 + 2HClu2192 CaClu2082 + 2Hu2082O.
n(HCl) : n(Ca(OH)u2082) = 2 : 1.
n(HCl) = 0.2 mol.