Supplemental Instruction: Chem177
Practice Exam 2
March 3, 2009
show all work and report answers in correct # of sig figs
1. Formic acid, HCHO2, is a weak electrolyte. What solute particles (only) are
present in an aqueous solution of this compound?
A. H+ and CHO2–
B. H+, CO, and OHC. HCHO2, H+, and CHO2–
D. HCHO2
2. An element that loses electrons is said to have been…
A. Redoxed
B. Oxidized
C. Reduced
D. Decomposed
3. What is an example of an extensive property?
A. heat capacity
B. density
C. specific heat
D. molarity
4. Which of the following solutions is the most acidic?
A. 0.5 M NH3
B. 0.3 M HNO3
C. 0.4 M HI
D. 0.7 M CH3OH
5. A chemical reaction that absorbs heat from the surroundings is said to be
__________ and has a __________ ΔH at constant pressure.
A. endothermic; positive
B. endothermic; negative
C. exothermic; negative
D. exothermic; positive
E. exothermic; neutral
6. Enthalpy is symbolized by
A. S
C. w
B. q
D. H
7. Which of the following is not a state function?
A. E
B. q
C. P
D. All are state functions
7. A compound consists of 75.69% C, 8.80% H, and 15.51% O by mass. Determine
the empirical formula as well as the molecular formula if the molar mass is
determined to be 412 g/mol.
*Assume 100 g
75.69g C x 1 mol/12.01 g = 6.30 mol C
8.80g H x 1 mol/1.01 g = 8.71 mol H
15.51g O x 1 mol/16.00 g = 0.97 mol O
*divide all by 0.97 to equal 6.5 mol C, 9 mol H, 1 mol O, times by 2
Emp. formula = C13H18O2
Mol. formula= C26H36O4
8.
Mg3N2(s) + 4H2SO4(aq)
3MgSO4(aq) + (NH4)2SO4(aq)
A. Balance the equation.
B. Calculate the number of grams of MgSO4 theoretically formed when 4.6 g
of Mg3N2 is reacted with 125 mL of 2.5M solution of H2SO4.
*4.6 g x 1 mol/100.9g = 0.046 mol Mg3N2 *limiting (from equation)
*2.5 M = x moles/0.125 L x = 0.3125 mol H2SO4
0.046 mol Mg3N2 x 3 mol MgSO4/1 mol Mg3N2 x 120.3 g/1 mol =
16.60 g MgSO4 (16 with sig figs)
C. If the actual yield of MgSO4 is 12.36 g, what is the percent yield?
12.36 g / 16.60 g = 75 % yield
D. How many moles of excess reactant remain?
0.046 mol Mg3N2 x 4 mol H2SO4/ 1 mol Mg3N2 = 0.184 mol H2SO4 used in rxn
0.3125 – 0.184 = 0.1285 mol (0.13 with sig figs)
9. Complete the following reaction, then write the balanced net ionic reaction
equation; show phases of all reactants and products.
2HC2H3O2(aq) + Ba(OH)2(aq)
2H20(l) + Ba(C2H3O2)2 (aq)
HC2H3O2(aq) + OH-
H20(l) + C2H3O2-
Identify the precipitant (if present): ________(none)______________________
Identify all spectator ions (if any present): ___Ba2+________________________
10. A stock solution of NaOH is 6.00M. How many mL of this solution is needed to
make 500 mL of 0.75M NaOH solution?
0.75 M NaOH = x moles/ 0.500 L
0.375 moles NaOH/x L = 6 M
x = 62.5 mL NaOH ( 60 mL with sig figs)
11. Assign oxidation numbers and decide what is oxidized and what is reduced.
PbS(s) + 4H2O2(aq)
oxidized:
sulfur in PbS
reduced:
oxygen in H2O2
PbSO4(s)
+
4H2O(l)
12. 4Fe(s) + 3O2(g)
2Fe2O3(s)
∆H = -1648.4 kJ
If 2.6 g of Fe is reacted in excess oxygen, how many kJ of heat is released?
2.6 g Fe x 1 mol/55.85 g x -1648.4 kJ/4 mol Fe = -19 kJ
13. If the specific heat of water is 4.18 J/g*K, how much heat (in J) is required to
raise the temperature of 72.0 grams of water by 12⁰C?
q = 72.0 g (4.18 J/g*K) ( 12⁰C)
q = 3600 J
14.
2C2H2(g) + 5O2(g)
4CO2(g) + 2H2O(l)
A. Using the standard enthalpies of formation, calculate the ∆H of the
reaction.
(4(-393.5) + 2(-285.8)) - (2(226.7)) = -2599 kJ
B. Is work done in this reaction? __yep___ If so, calculate amount of work and
the ∆E.
w = - 1 atm ((4-7)(22.4L) = 67.16 L*atm
w = 67.16 x 101.3 J/L*atm = 6797 J
∆E = -2599 + 6.79 kJ = -2593 kJ
15. Find the concentration of each ion present when 16 mL of 1.0M KNO3 is mixed
with 22 mL of 0.75M K2SO4.
1.0 M KNO3 = x mol/0.016 L = 0.016 mol KNO3
0.75 M K2SO4 = x mol/0.022 L = 0.0165 mol K2SO4
K+: 0.016 + 2(0.0165) = 0.049 mol/0.038 L = 1.29 M
NO3-: 0.016 mol/0.038 L = 0.42 M
SO42-: 0.0165 mol/0.038 L = 0.43 M