Steps for solving Stoichiometric Problems Involving Solution

1
Steps for solving Stoichiometric Problems
Involving Solution
Step 1 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 1.50 L of a 0.100 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)
0.150 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
Step 4 Calculate the moles of Cl- required We
have 0.150 mol of Ag ions and , because one Ag
ion reacts with one Cl- ion, we need 0.150 mol of
Cl- Ag(aq) Cl-(aq) ?? AgCl(s) So 0.150
mol of AgCl will be formed.
Step 5 Convert to grams of NaCl required To
produce 0.150 mol Cl- , we need 0.150 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 0.100
M HCl solution needed to neutralize25.0 mL of a
0.350 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)
0.00875 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
Step 4 Calculate the moles of H required We
have 0.00875 mol of OH- ions and , because one
OH- ions reacts with one H ions, we need 0.00875
mol of H H(aq) OH-(aq) ?? H2O(l) So
0.00875 mol of H2O will be formed.
Step 5 Calculate the volume of 0.100M HCl
required To produce 0.150 mol Cl- , we need 0.150
mol of NaCl. We calculate the mass ofNaCl
required as follows. Volume n / molarity
(0.00875 mol)/(0.100molL-1)
0.0875 L
87.5 mL
Therefore, 87.5mL of 0.100M HCl is required to
neutralize 25.0 mL of 0.350 M NaOH
8
Normality