CHEM 100
Chemistry I
Exam II—Optional Make-up
30-September-2005
Answers
1) A sample of dolomitic limestone containing only CaCO3 and MgCO3 was
analyzed.
a) When a 0.2800 g sample of this limestone was decomposed by heating, 75.0
mL of CO2 at 750 mm Hg and 20°C were evolved. How many grams of CO2 were
produced?
Applying the ideal gas law, rearragned, gives us: n =
 750mmHg
(0.075 L )
760mmHg / atm 

= 3.08 x 10-3 moles, which, at 44.01 g/mol for
(0.08206 L ⋅ atm / K ⋅ mol )(293.15K )
CO2, give us 0.136 g of CO2.
b) Write equations for the decomposition of both carbonates described
above.
CaCO3 → CaO + CO2
MgCO3 → MgO + CO2
c) It was also determined separately that the initial sample contained 44.8 mg
of calcium. What percent of the limestone by mass was CaCO3?
The number of moles of Ca = number of moles of CaCO3, so: number of moles of
0.0448 g
= 1.12 x 10-3 moles so mass of CaCO3 =
Ca=number of moles of CaCO3 =
40.08 g / mol
0.1121g
(0.00112 moles)(100.089 g/mol) = 0.1121 g CaCO3 and percent CaCO3 =
x
0.2800 g
100 = 40.0%.
d) How many grams of the magnesium-containing product were present in
the sample in (a) after it had been heated.
If the initial sample had 0.112 g CaCO3, and massed 0.2800 g, then the mass of
MgCO3 = (0.2800-0.1121) = 0.1679 g MgCO3. All the Mg in the magnesium-containing
product (= MgO) came from this magnesium, so the number of moles is equal. Moles
0.1679 g
MgCO3 = moles MgO =
= 1.99 x 10-3 moles. This number of moles of
84.314 g / mol
MgO has a mass of: (1.99 x 10-3 moles)(40.3044 g/mol) = 0.0802 g
2) Magnesium metal reacts with aqueous HCl to yield H2 gas:
Mg(s) + 2 HCl(aq) -----> MgCl2(aq) + H2(g)
The gas that forms is found to have a volume of 3.557 L at 25°C and a
pressure of 747 mm Hg. Assuming that the gas is saturated with water vapor at a
partial pressure of 23.8 mm Hg, determine how many grams of magnesium metal
were used in the reaction.
The pressure of the hydrogen gas is (747 – 23.8) = 723.2 mm Hg = 0.952 atm.
Now we can determine the number of moles of H2, and from that, and the reaction
stoichiometry, the number of moles, and finally, grams, of Mg.
Moles H2, n =
(0.952atm )(3.557 L )
(0.08206 L ⋅ atm / K ⋅ mol )(298.15K )
= 0.138 moles. This is also
the number of moles of Mg, so the grams Mg = (0.138 moles)(24.305 g/mol) = 3.364 g
Mg.
3) Balance the following redox reactions by the half-reaction method. All species are
in aqueous solution unless otherwise stated.
a) Cl- + Sn(s) +NO3- → SnCl62- + NO2(g)
(acid)
6Cl- + Sn → SnCl62- + 4eNO3- + 2H+ + e- → NO2 + H2O
Multiply the second one by 4, add, and simplify.
6Cl- + Sn + 4NO3- + 8H+→ SnCl62- + 4NO2 + 4H2O
b) Ba
2+
(aq)
+ 2OH-(aq) + H2O2(aq) + ClO2(aq) → Ba(ClO2)2(s) + H2O(l) + O2(g)
Ba2+ + 2ClO2 + 2e- → Ba(ClO2)2
H2O2 → O2 + 2H+ + 2eAdd, and simplify to get:
Ba2+ + 2ClO2 + H2O2 → Ba(ClO2)2 + O2 + 2H+
but we’re in base, so add 2 OH- to both sides to get, finally:
Ba2+ + 2ClO2 + H2O2 + 2OH- → Ba(ClO2)2 + O2 + 2H2O.
c) NO3- + Zn(s) → Zn(OH)42- + NH3
(base)
Zn + 4H2O → Zn(OH)42- + 4H+ + 2eNO3- + 9H+ + 8e- → NH3 + 3H2O
Multiply first equation by 4, add, and simply:
4Zn + NO3- + 13H2O → 4Zn(OH)42- + NH3 + 7H+
but we are in base, so add 7OH- to both sides, and further simpify to get:
4Zn + NO3- + 6H2O + 7OH- → 4Zn(OH)42- + NH3.
4) A person’s blood alcohol (C2H5OH) level can be determined by titrating a sample
of blood plasma (which is where the alcohol ends up) with a potassium dichromate
solution. The balanced equation is:
16 H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2 CO2(g) + 11 H2O(l)
If 35.46 mL of 0.04961 M Cr2O72- is required to titrate 25.00 g of plasma, what is
the mass percent of alcohol in the blood?
Here is the balanced equation again, just so we can fill in important information:
16 H+(aq) + 2Cr2O72-(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2 CO2(g) + 11 H2O(l)
Mm
Molarity
Volume
Moles
Grams
46.0694
0.04961
35.46 mL
1.76 x 10-3
Percent alcohol =
states.)
8.80 x 10-4 (from stoichiometry)
0.0405 g
0.0405 g
x 100 = 0.162 % (which is twice the legal limit in many
25.00 g