AP Chemistry Name
1 ▪ Describing Matter Period Date
1.3 PROBLEMS – STOICHIOMETRY
1. A noncarbonated soft drink contains an unknown amount of citric acid, H 3C6H5O7. If 100. mL of the soft
drink requires 33.51 mL of 0.0102 M NaOH to neutralize the citric acid completely, how many grams of
citric acid does the soft drink contain per 100. mL? The reaction of citric acid and NaOH is
H3C6H5O7 (aq) + 3 NaOH (aq) → Na3C6H5O7 (aq) + 3 H2O (ℓ)
(Ch 5, #86)
Molar Mass of H3C6H5O7 = 6(12.01) + 8(1.008) + 7(16.00) = 192.12 g/mol
33.51 mL = 0.0219 g H3C6H5O7
2. Vitamin C is the simple compound C6H8O6. Besides being an acid, it is also a reducing agent. One method
for determining the amount of vitamin C in a sample is there to titrate it with a solution of bromine, Br 2, an
oxidizing agent.
C6H8O6 (aq) + Br2 (aq) → 2 HBr (aq) + C6H6O6 (aq)
Suppose a 1.00 g “chewable” vitamin C tablet requires 27.85 mL of 0.102 M Br 2 for titration to the
equivalence point. How many grams of vitamin C are in the tablet? (Ch, 5, #90)
Molar Mass of C6H8O6 = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 g/mol
27.85 mL = 0.500 g C6H8O6
3. Sodium thiosulfate, Na2S2O3, is used as a “fixer” in black-and-white photography. Assume you have a bottle
of sodium thiosulfate and want to determine its purity. The thiosulfate ion can be oxidized with I 2 according
to the equation
I2 (aq) + 2 S2O32– (aq) → 2 I– (aq) + S4O62– (aq)
If you use 40.21 mL of 0.246 M I2 in a titration, what is the weight percent of Na2S2O3 in a 3.232-g sample of
impure material? (Ch 5, #100)
Molar Mass of Na2S2O3 = 2(22.99) + 2(32.06) + 3(16.00) = 158.10 g/mol
40.21 mL = 3.13 g Na2S2O3
× 100% = 96.8% Na2S2O3
4. Suppose you mix 25.0 mL of 0.234 M FeCl3 solution with 42.5 mL of 0.453 M NaOH. What is the maximum
mass, in grams, of Fe(OH)3 that precipitates? Which reactant is in excess? What is the molar concentration
of the excess reactant remaining in solution after the maximum mass of Fe(OH) 3 has been precipitated?
(Ch. 5, #108)
FeCl3 (aq) + 3 NaOH (aq) → Fe(OH)3 (s) + 3 NaCl (aq)
Molar Mass of Fe(OH)3 = 1(55.85) + 3(16.00) + 3(1.008) = 106.87 g/mol
Method 1:
25.0 mL = 0.625 g Fe(OH)3
42.5 mL = 0.686 g Fe(OH)3
0.625 g Fe(OH)3 precipitates. FeCl3 is the limiting reactant, and NaOH is the excess reactant.
42.5 mL = 0.0193 mol NaOH initially present
25.0 mL = 0.0176 mol NaOH consumed
0.0193 mol – 0.0176 mol = 0.0017 mol NaOH remain
[NaOH] = = 0.025 M NaOH in final mixture
Method 2:
25.0 mL = 0.00585 mol FeCl3 initially present
42.5 mL = 0.0193 mol NaOH initially present
moles FeCl3 (aq) + 3 NaOH (aq) → Fe(OH)3 (s) + 3 NaCl (aq)
Initial 0.00585 0.0193 0 0
Change –0.00585 – 3(0.00585) + 0.00585 + 3(0.00585)
Final 0 0.00175 0.00585 0.0176
Mass of Fe(OH)3 (s) formed = 0.00585 g = 0.625 g Fe(OH)
[NaOH] = = 0.025 M NaOH in final mixture
5. You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a sample of
an alloy in acid, an excess of KI is added, and the Cu 2+ and I– ions undergo the reaction
2 Cu2+ (aq) + 5 I– (aq) → 2 CuI (s) + I3– (aq)
The liberated I3– is titrated with sodium thiosulfate according to the equation
I3– (aq) + 2 S2O32– (aq) → S4O62– (aq) + 3 I–
If 26.32 mL of 0.101 M Na2S2O3 is required for titration to the equivalence point, what is the weight percent
of Cu in 0.251 g of the alloy? (Ch 5., #109)
26.32 mL = 0.169 g Cu
× 100% = 67.3% Cu
6. Iron reacts with hydrochloric acid to produce iron(II) chloride and hydrogen gas:
Fe (s) + 2 HCl (aq) → FeCl2 (aq) + H2 (g)
The H2 gas from the reaction of 2.2 g of iron with excess acid is collected in a 10.0 L flask at 25°C. What is
the pressure of the H2 gas in this flask? (Ch. 12 #44)
2.2 g Fe = 0.039 mol H2
P = = = 0.096 atm
7. Sodium azide, the explosive compound in car air bags, decomposes according to the equation
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
What mass of sodium azide is required to provide the nitrogen needed to inflate a 25.0 L bag to a pressure of
1.3 atm and 25°C? (Ch. 12, #46)
Molar Mass of NaN3 = 1(22.99) + 3(14.01) = 65.02 g/mol
nN2 = = = 1.3 mol N2
1.3 mol N2 = 58 g NaN3
AP Chemistry 2008B #3
A 0.150 g sample of solid lead(II) nitrate is added to 125 mL of 0.100 M sodium iodide solution. Assume no
change in volume of the solution. The chemical reaction that takes place is represented by the following equation.
Pb(NO3)2(s) + 2 NaI(aq) → PbI2(s) + 2 NaNO3(aq)
(a) List an appropriate observation that provides evidence of a chemical reaction between the two compounds.
Formation of yellow PbI2 precipitate
(b) Calculate the number of moles of each reactant.
Molar mass of Pb(NO3)2 = 1(207.2) + 2(14.01) + 6(16.00) = 331.2 g/mol
0.150 g Pb(NO3)2 = 0.000453 mol Pb(NO3)2
125 mL = 0.0125 mol NaI
(c) Identify the limiting reactant. Show calculations to support your identification.
0.000453 mol Pb(NO3)2= 0.000906 mol NaI needed
With 0.000453 mol Pb(NO3)2, 0.00906 mol NaI is needed to react completely. Since we have 0.0125
mol NaI, NaI is the excess reactant, and Pb(NO3)2 is the limiting reactant.
(d) Calculate the molar concentration of NO 3−(aq) in the mixture after the reaction is complete.
Since NO3– is soluble and is therefore a spectator ion, the moles of NO 3– after the end of the reaction
is equal to the moles of NO3– from the Pb(NO3)2 originally used.
[NO3–] = = 0.00725 M NO3–
(e) Circle the diagram below that best represents the results after the mixture reacts as completely as
possible. Explain the reasoning used in making your choice.
Since Pb(NO3)2 is the limiting
reactant, all of the Pb2+ will be a
part of the precipitate, and the
remaining NO3– will remain in
solution. Since NaI is the excess
reactant, not all of the I– will
precipitate out, so some of it will
remain in solution. Finally, since
NaI is in excess and Pb(NO3)2 is
limiting, there will be more free Na+
than NO3– in solution.
