1. Steps for solving Stoichiometric Problems Involving Solution
Write the balanced equation for the reaction.
Step 2
Calculate the moles of reactants
Step 3
Calculate the limiting reactant
Step 4
Calculate the moles of other reactants or products
Step 5
Convert to grams or other units, if required.

2. Practice writing net ionic equations for these reactions.
A net inoic equation only shows the components that are directly involved in the reaction.
K2CrO4 (aq) + Ba(NO3)2 (aq)  BaCrO4 (s) + 2KNO3 (aq)
CrO4- (aq) + Ba2+ (aq)  BaCrO4 (s)
Na2SO4 (aq) + Pb(NO3)2 (aq)  PbSO4 (s) + 2NaNO3 (aq)
SO42- (aq) + Pb2+ (aq)  PbSO4 (s)
Na2S (aq) + NiCl2 (aq)  NiS (s) + 2NaCl (aq)
S2- (aq) + Ni2+ (aq)  NiS (s)
NaOH (aq) + FeCl3 (aq)  Fe(OH)3 (s) + 2NaCl (aq)
3OH- (aq) + Fe3+ (aq)  Fe(OH)3 (s)

3. Problem 1:. Calculate the mass of solid NaCl that must be added to 1
Problem 1: Calculate the mass of solid NaCl that must be added to 1.50 L of a M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl. Calculate the mass of AgCl formed.
Step 1: Write a balance equation
Ag+(aq) + Cl-(aq)  AgCl(s)
Step 2: Calculate the moles of reactants
n = M  V = (0.100mol L-1)(1.50L)
= mol Ag+
Step 3: Determine which reactant is limiting
We are adding just enough Cl- to react with the Ag+ present. So the Ag+ is the limiting reactant

4. Ag+(aq) + Cl-(aq)  AgCl(s)
Step 4: Calculate the moles of Cl- required
We have mol of Ag+ ions and , because one Ag+ ion reacts with one Cl- ion, we need mol of Cl- :
Ag+(aq) + Cl-(aq)  AgCl(s)
So mol of AgCl will be formed.
Step 5: Convert to grams of NaCl required
To produce mol Cl- , we need mol of NaCl. We calculate the mass ofNaCl required as follows.
Mass = n  Molar mass
The mass of AgCl formed is
= (0.150 mol)(58.4gmol-1)
= 8.76 g NaCl
Mass = n  Molar mass
= (0.150 mol)(143.3gmol-1)
= 21.5 g AgCl

5. Neutralization reactions
An acid-base reaction is often called a neutralization reaction. This is because when you add just enough strong base to react exactly with a strong acid in a solution, we say the acid has been neutralized.
One product of a neutralization reaction is always water.
Acid + a base = water and a salt
HCl (aq) + KOH (aq)  H20(l) + KCl (aq)
The net reaction is: H+ (aq) + OH- (aq)  H20(l)

6. Problem 1:. Calculate the volume of a 0
Problem 1: Calculate the volume of a M HCl solution needed to neutralize25.0 mL of a M NaOH solution.
Step 1: Write a balance equation
H+(aq) + OH-(aq)  H2O(l)
Step 2: Calculate the moles of reactants
n = M  V = (0.350mol L-1)(0.025L)
= mol OH-
Step 3: Determine which reactant is limiting
We are adding just enough H+ ions to react exactly with the OH- ions present. So the OH- is the limiting reactant

7. H+(aq) + OH-(aq)  H2O(l)
Step 4: Calculate the moles of H+ required
We have mol of OH- ions and , because one OH- ions reacts with one H+ ions, we need mol of H+:
H+(aq) + OH-(aq)  H2O(l)
So mol of H2O will be formed.
Step 5: Calculate the volume of 0.100M HCl required
To produce mol Cl- , we need mol of NaCl. We calculate the mass ofNaCl required as follows.
Volume = n / molarity
= ( mol)/(0.100molL-1)
= L
= 87.5 mL
Therefore, 87.5mL of 0.100M HCl is required to neutralize 25.0 mL of M NaOH