Stoichiometry – Summer Review for AP Chemistry
Name: ___ANSWER KEY___________
NOTE: Answer Key Needs to be checked – it is possible that there COULD be some errors below:
1. A sample of glucose, C6H12O6 contains 1.250 X 1021 atoms of carbon.
1A. How many atoms of hydrogen does it contain?
Since the ratio between Carbon and Hydrogen is 1:2 (6:12), then there will be twice as many
hydrogen atoms as there are carbon atoms = 2.500 x 1021 atoms of hydrogen
1B. How many molecules of glucose does it contain?
Since glucose needs six atoms of carbon to make one molecule of glucose:
1 molecule of glucose
21
1.250 X 10 atoms of carbon x ------------------------------- = 2.083 x 1020 glucose molecule
6 atoms of Carbon
1C. How many moles of glucose does it contain? (Need to use Avogadro’s Number)
1 mol glucose
2.083 x 1020 glucose molecules x ------------------------------------------ = 0.0003459 mol glucose
6.022 x 1023 molecules glucose
1D. What is the mass of this sample in grams?
MM: C6H12O6 = 6(C) + 12 (H) + 6 (O) = 6(12.01g) + 12 (1.01g) + 6 (16.00g) = 180.18 g/mol
180.18 g glucose
0.0003459 mol glucose x ---------------------------------------- =
1 mol glucose
0.06232 g glucose
2. Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which
means that a certain number of water molecules are included in the solid structure. Its
formula can be written as Na2CO3 • xH2O, where x is the number of moles of H2O per mole of
Na2CO3. When a 2.558 g sample of washing soda is heated at 250C, all the water of hydration is
lost, leaving 0.948 g of Na2CO3. What is the value of x?
NOTE: We did not discuss this in general chemistry – HOWEVER, there are going to be MANY
times when you will be REQUIRED to learn things on-your-own. In AP Chemistry, there can NOT
be any excuses, so hopefully you figured this one out on-your-own.
Difference in mass: 2.558 g (mass of hydrate) - 0.948g (mass of Na2CO3) = 1.610 g H2O
MM: H2O = 2(H) + 1 (O) = 2(1.01g) + 1(16.00g) = 18.02 g/mol H2O
MM: Na2CO3 = 2(22.99g) + 1(12.01g) + 3(16.00g) = 105.99 g/mol Na2CO3
1 mol Na2CO3
0.948 g Na2CO3 x ----------------------- = 0.00894 mol Na2CO3
105.99g Na2CO3
1 mol H2O
1.610 g H2C x ----------------------- = 0.08934 mol H2O
18.02g H2O
Note the 1:10 ratio between
_.08934 mol H2O__ = 10 mol H2O__
.00894 mol Na2CO3
1 mol Na2CO3
0.00894 mol Na2CO3 and 0.08934 mol H2O
Therefore, the value of “x” is 10  written as Na2CO3 • 10 H2O (the “•” is used for hydrates)
Note: For every one mole of Na2CO3 there is ten moles of H2O
3. Given this equation:
2 KI + Pb(NO3)2  PbI2 + 2 KNO3
calculate mass of PbI2 produced by reacting of 30.0 g KI with excess Pb(NO3)2
MM: PbI2 = 1(Pb) + 2(I) = 1(207.2g) + 2 (126.90g) = 461.0 g/mol PbI2
MM: KI = 1(K) + 1(I) = 1(39.10g) + 1 (126.90g) = 166.00 g/mo KI
1 mol KI
1 mol PbI2
461.0 g PbI2
30.0 g KI x ------------------ x ----------------- x ----------------- = 41.6 g PbI2
166.00 g KI
2 mol KI
1 mol PbI2
4. The reaction of 2.5g of aluminum with 2.5g of oxygen (actually) produces 3.5g of aluminum
oxide. Calculate the percent yield of aluminum oxide.
4Al + 3O2  2Al2O3 Need to find the Limiting Reactant to determine Theoretical Yield
2.5 g
2.5g
4.71 g (NOTE: 2.5g + 2.5g does not equal 4.71 g – excess is present)
MM: Al2O3 = 2(Al) + 3(O) = 2(26.85g) + 3 (16.00g) = 101.70 g/mol Al2O3
MM: O2 = 2(O) = 2(16.00g) = 32.00 g/mol O2
MM: Al = 26.98 g/mol Al
1 mol Al
2 mol Al2O3
101.70 g Al2O3
2.5 g Al x ------------------ x ----------------- x ----------------------------- = 4.71 g Al2O3 Al = Limiting
26.98 g Al
4 mol Al
1 mol Al2O3
Theoretical Yield Reactant
1 mol O2
2 mol Al2O3
101.70 g Al2O3
2.5 g O2 x ------------------ x ----------------- x ----------------------------- = 5.29 g Al2O3 O2 = Excess
32.00 g O2
3 mol O2
1 mol Al2O3
Reactant
Actual yield
3.5 g Al2O3
Percent Yield = --------------------------- x 100% = ------------------ x 100 % = 74 % yield Al2O3
Theoretical Yield
4.7 g Al2O3
NOT Required, but just practice  Calculate excess O2
1 mol Al
3 mol O2
32.00 g O2
2.5 g Al x ------------------ x ----------------- x ------------------- = 2.22 g O2 = needed oxygen
26.98 g Al
4 mol Al
1 mol O2
Excess =
Have – Need
=
2.5 g O2 – 2.2 g O2 = 0.3 g O2 in excess
(Self Check) O2 + Al = Al2O3
2.2g + 2.5g = 4.7g
5. Disulfur dichloride, which has a revolting smell, can be prepared by directly combining S8 and
Cl2, but it can also be made by the following reaction:
3 SCl2(l) + 4 NaF(s)  SF4(g) + S2Cl2(l) + 4 NaCl(s)
5.23 g
excess
__________g
Assume you begin with 5.23 g of SCl2 and excess NaF. What is the theoretical yield of S2Cl2 ? If only
1.19 g of S2Cl2 is obtained, what is the percent yield of the compound?
MM: SCl2 = 1(S) + 2(Cl) = 1(32.07g) + 2(35.45g) = 102.97 g/mol SCl2
MM: S2Cl2 = 2(S) + 2(Cl) = 2(32.07g) + 2(35.45g) = 135.04 g/mol S2Cl2
1 mol SCl2
1 mol S2Cl2
135.04 g S2Cl2
5.23 g SCl2 x ------------------ x ----------------- x ----------------------------- = 2.287 g S2Cl2
102.97 g Al
3 mol SCl2
1 mol S2Cl2
Theoretical Yield
Actual yield
1.19 g S2Cl2
Percent Yield = --------------------------- x 100% = ------------------ x 100 % = 52.03 % yield S2Cl2
Theoretical Yield
2.287 g S2Cl2
6. Silver nitrate and aluminum chloride react with each other by exchanging anions.
AgNO3 + AlCl3  Al (NO3)3 + AgCl
_3_AgNO3 + _1_AlCl3  _1_Al (NO3)3 + _3_AgCl
4.22 g
7.73 g
g
.
What mass of AgCl is produced when 4.22g of silver nitrate reacts with 7.73g of aluminum
chloride? Balance the equation first. NOTE: Need to find limiting reactant.
MM: AgNO3 = 1(107.87g) + 1(14.01g) + 3(16.00g) = 169.88 g/mol AgNO3
MM: AlCl3 = 1(26.98g) + 3(35.45g) = 133.33 g/mol AlCl3
MM: AgCl = 1(107.87g) + 1(35.45g) = 143.32 g/mol AgCl
1 mol AgNO3
3 mol AgCl
143.32 g AgCl
4.22 g AgNO3 x ------------------- x ----------------- x ---------------------- = 3.560 g AgCl Limiting
169.88 g AgNO3
3 mol AgNO3
1 mol AgCl
Reactant
1 mol AlCl3
3 mol AgCl
143.32 g AgCl
7.73 g AlCl3 x ------------------ x ----------------- x ------------------------ = 24.92 g AgCl Excess
133.33 g AlCl3
1 mol AlCl3
1 mol AgCl
Reactant
The maximum amount of AgCl possible (based on limiting reactant) = 3.560 g AgCl
NOT Required, but just practice  Calculate excess AlCl3
1 mol AgNO3
1 mol AlCl3
133.33 g AlCl3
4.22 g AgNO3 x ------------------ x ----------------- x ------------------ = 1.104 g AlCl3 = needed AlCl3
169.88 g AgNO3 3 mol AgNO3
1 mol AlCl3
Excess =
Have
–
Needed
=
7.73 g AlCl3 – 1.10 g AlCl3 = 6.93 g AlCl3 in excess
NOTE: The mass of Al (NO3)3 produced can be determined by using the law of conservation of mass:
(Mass of Reactants) = (Mass of Products) (4.22 AgNO3 g + 1.10 AlCl3g) = ( 1.76 g Al (NO3)3 + 3.56 g AgCl)
Aqueous Reactions and Solution Stoichiometry
7. Write balanced net ionic equations for the reactions that occur in each of the following cases.
Identify the spectator ion or ions in each reaction. NOTE: Phase labels are REQUIRED
A. _1_Cr2(SO4)3 (aq) + _3_ (NH4)2CO3 (aq)  _1_ Cr2(CO3)3 (s) + _3_ (NH4)2SO4 (aq) CO32- = “I”
2 Cr 3+(aq) + 3 SO42- (aq) + 6 NH41+(aq) + 3 CO32-(aq)  _1_ Cr2(CO3)3 (s) + 6 NH41+(aq) + 3 SO42- (aq)
2 Cr 3+(aq) + 3 CO32-(aq)  _1_ Cr2(CO3)3 (s)
Spectator ions = 6 NH41+(aq) + 3 SO42- (aq)
B. _1_Ba(NO3)2 (aq) + _1_K2SO4 (aq)  _1_ BaSO4 (s) + _2_KNO3 (aq)
SO42- = “S” except CBS/APH
1 Ba2+(aq) + 2 NO31- (aq) + 2 K1+ (aq) + 1 SO42- (aq)  _1_ BaSO4 (s) + 2 K1+ (aq) + 2 NO31- (aq)
1 Ba2+(aq) + 1 SO42- (aq)  _1_ BaSO4 (s)
Spectator ions = 2 NO31- (aq) + 2 K1+ (aq)
C. _1_Fe(NO3)2 (aq) + _2_KOH(aq)  _2_ KNO3 (aq) + _1_Fe(OH)2 (S)
OH1- = “I” except CBS, K1+ = “S”
1 Fe2+(aq) + 2 NO31- (aq) + 2 K1+ (aq) + 2 OH1- (aq)  2 NO31- (aq) + 2 K1+ (aq) + _1_Fe(OH)2 (S)
1 Fe2+(aq) + 2 OH1- (aq)  _1_Fe(OH)2 (S)
Spectator ions = 2 NO31- (aq) + 2 K1+ (aq)
NOTE: a “net ionic” equation is comprised of aqueous ions for the reactants and the products are NOT ions
as the reactant ions chemical combine to form a solid precipitate, a gas, and even some liquids. If
all of the ions are spectators (aqueous on both sides), then no reaction has occurred = NR. This is
why it is important to know solubility rules in terms of forming “insoluble” compounds.
8. A. Calculate the molarity of a solution that contains 0.0345 mol NH4Cl in exactly 400 mL of
solutions. Molarity = moles of solute/Liter of solution (Need to convert mL to L)
M = 0.0345 mol NH4Cl / .400 L solution = 0.08625 M NH4Cl(aq)
NOTE: Exactly 400 mL states that 400 has unlimited sig figs.
B. How many moles of HNO3 are present in 35.0 mL of a 2.20 M solution of nitric acid?
moles solute = Molarity * liters solution = 2.20 M * 0.0350 L =
0.0770 moles of HNO3
C. How many milliliters of 1.50 M KOH solution are needed to provide 0.125 mol of KOH.
Liters solution = moles of solute/Molarity = 0.125 mol / 1.50 M = 0.08333 L of 1.50 M KOH
Remember to convert to milliliters as the question REQUIRES = 83.33 mL of 1.50 M
KOH
9. A. Which will have the highest concentration of potassium ion: 0.20M KCl, 0.15 M K2CrO4, or
0.080M K3PO4?
KCl = 1 mol K+ per 1 mol KCl
0.20M KCl * 1mol K+/1 mol KCl = 0.20M K+
K2CrO4 = 2 mol K+ per 1 mol K2CrO4
0.15M K2CrO4 * 2mol K+/1 mol K2CrO4 = 0.30M K+
K3PO4 = 3 mol K+ per 1 mol K3PO4
0.080M K3PO4 * 3mol K+/1 mol K3PO4 =0.24M K+
The 0.15 M
K2CrO4 will have the highest potassium ion
concentration at 0.30 M K+
NOTE: My final answer was very clear and easy to understand AND I provided mathematical
evidence to support my answer. My answer is also in units of concentration M.
9. B. Which will contain the greater number of moles of potassium ion: 30.0 mL of 0.15 M K2CrO4,
or 25.0 mL of 0.080 M K3PO4? Molarity = moles of solute/Liter of solution
NOTE: First need to determine the number of moles of each compound, then determine the number
of moles of potassium ion using the molar ratio of the compound. (Convert mL to L)
moles solute = Molarity * Liters solution
0.15 M * 0.0300 L = 0.0045 moles of K2CrO4 * 2mol K+/1 mol K2CrO4 = 0.0090 moles of K+
0.080 M * 0.0250 L = 0.0020 moles of K3PO4 * 3mol K+/1 mol K3PO4 = 0.0060 moles of K+
The 30.0 mL of 0.15 M K2CrO4 has the greater number of moles of K+ at 0.0090 mol K+
NOTE: My final answer was very clear and easy to understand AND I provided mathematical
evidence to support my answer. My answer is also in units of moles.