Practice exam #2 key
10-10-06
1. The temperature of an ideal gas in a sealed glass container originally at
760 mmHg pressure and 25°C is lowered to 220 K. Calculate the new
pressure of the gas in atm.
0.738 atm
P1V1/n1T1 = P2V2/n2T2
2. Based on solubility rules, categorize the following compounds as soluble
or insoluble in water at 25°C. Refer to Table 4.2 (p. 97) if necessary.
A. Na2CO3
soluble
insoluble
B. CaCO3
soluble
insoluble
C. BaSO4
soluble
insoluble
D. (NH4)2S
soluble
insoluble
E. SrSO4
soluble
insoluble
3. (Multiple Choice) Which of the following does not represent an oxidationreduction (redox) reaction?
A.
B.
C.
D.
E.
3Al + 6HCl → 3H2 + AlCl3
2H2O → 2H2 + O2
2NaCl + Pb(NO3)2 → PbCl2 + 3NaNO3
2NaI + Br2 → 2NaBr + I2
Cu(NO3)2 + Zn → Zn(NO3)2 + Cu
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Practice exam #2 key
10-10-06
4. Circle the oxidizing agent in each of the following chemical reactions:
A. Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)
B. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
C. 2KBr(aq) + Cl2(g) → 2KCl(aq) + Br2(l)
D. 2H2(g) + O2(g) → 2H2O(l)
5. What mass of Li3PO4 is needed to prepare 500 mL of a solution having a
lithium ion concentration (Li+) of 0.175 M?
3.38 g
Note: 3 mol Li+ per 1 mol Li3PO4
Find mol Li+ ion from MV=# mol
Calculate # mol Li3PO4 → # grams
6. 0.820 mol of hydrogen gas (H2(g)) has a volume of 2.00 L at a certain
temperature and pressure. What is the volume of 0.125 mol of this gas at
the same temperature and pressure?
0.305 L
P1V1/n1T1 = P2V2/n2T2
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Practice exam #2 key
10-10-06
7. Determine the molar mass of chloroform gas if a sample weighing 0.389 g
is collected in a volume of 102 cm3 at 97°C. The pressure of the
chloroform is 728 mmHg.
121 g/mol
Note: find # moles using PV=nRT (watch units!). Molar mass = #g/#mol.
8. When active metals such as magnesium are immersed in acid solution,
hydrogen gas is evolved (the other product is magnesium chloride,
MgCl2). Calculate the volume of hydrogen gas (H2(g)) at 30.1°C and 0.85
atm that can be formed when 275 mL of 0.725 M HCl solution reacts with
excess Mg(s) to give hydrogen gas and aqueous magnesium chloride.
2.92 L
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
Find # moles of HCl reacted (limiting reagent)…find # moles H2 evolved. Use
PV=nRT to find V.
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Practice exam #2 key
10-10-06
9. When 38.0 mL of 0.125 M H2SO4 is added to 100 mL of a solution of PbI2,
a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution,
dried, and weighed. If the recovered PbSO4 is found to have a mass of
0.0471 g, what was the concentration of lead ions (Pb2+) in the original
solution?
1.6 x 10-3 M
H2SO4 + PbI2 → PbSO4 + 2HI
We recovered 1.6 x 10-4 mol PbSO4(s). Yield 1 mol PbSO4 for every 1 mol PbI2
reacted, so you started with 1.6 x 10-4 mol PbI2 in 0.100 L: divide to get mol/L
(M). The volume and molarity of H2SO4 tells you that indeed PbI2 was the
limiting reagent.
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Practice exam #2 key
10-10-06
10. A volume of 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0
mL of an H2SO4 solution.
A. Write a molecular equation describing the reaction.
H2SO4 + 2NaOH → 2H2O + Na2SO4
B. Write a net ionic equation describing the reaction.
H+ + OH- → H2O
C. What was the concentration of sulfuric acid (H2SO4) in the original
solution?
5.3 x 10-2 M
(Mbase)(Vbase) = (Macid)(Vacid)
Mbase = 0.1510 M
Vbase = 0.03462
Macid = ?
Vacid = 0.0500 L
Solve for Macid
Macid = 2 x MH2SO4
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Practice exam #2 key
10-10-06
11. Approximately 80% of the mass of a human body consists of aqueous
solutions (blood, cerebral/spinal fluid, etc.). These solutions are salt solutions
with a concentration of 0.21 M NaCl and have a density of 1 g/mL. How
many grams of NaCl are present the body of an adult who weighs 82.0 kg?
8.1 x 102 g NaCl
82 kg x 80% → the body holds 65.6 kg of salt solution
using density, the salt solution has a volume of 65.6 L
65.6 L salt x 0.21mol/L = 13.8 mol NaCl x molar mass = 806.5 g
12. There is a power plant in Portland, Oregon that is very concerned about
global warming. This plant takes all of its exhaust gases from its boilers and
recycles the CO2 using the Solvay process to make sodium hydrogen
carbonate (NaHCO3). The reaction is shown below:
NH3(g) + H2O(l) + CO2(g) + NaCl(aq) → NaHCO3(aq) + NH4Cl(aq)
How many liters of each NH3 and CO2 (both at STP) are needed to make
3.00 kg of sodium hydrogen carbonate?
7.99 x 102 L of NH3 and 7.99 x 102 L of CO2
3000 g NaHCO3 = 35.7 mol NaHCO3
1 mol NH3 reacted yields 1 mol NaHCO3, so you need 35.7 mol NH3
PV = nRT
VNH3 = 799 L
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