name answer key spring 2010 exam ii recitation ta grading ......1 mol nitroglycerin = 619 g (3 sig...
Post on 02-Mar-2021
2 Views
Preview:
TRANSCRIPT
Dr. Joe Burnett Spring 2010 This exam consists of 13 questions on 9 pages
Chem 177 Exam II
March 3, 2010
Name______________________ Recitation TA ______________ Recitation Section ___________
Grading Teaching Assistants and Recitation Sections
Page Points Score Name Sections Time 2 3 4 5 6 7
TOTAL
24 pts
14 pts
24 pts
14 pts
16 pts
12 pts
104 pts
________
________
________
________
________
________
Stacey Althaus
Kyle Marchuk
Kristopher McKee
Justin Smith
Anthony Stender
Nishad Thamban Chandrika
Songchen Xu
5
1, 10
11, 12
8
2, 3
4, 6
7, 9
10:00
8:00, 1:10
2:10, 1:10
12:10
8:00, 9:00
9:00, 11:00
11:00, 12:10
Questions are written on both sides of each page. There is a periodic table and an equation on the last page. You may remove this page for scratch paper. Show all work. Answers should include appropriate units and contain the correct number of significant figures. The last day to drop Chem 177 and add Chem 50 is Thursday, March 4, at 4:00 PM. See your advisor to obtain the drop/add form and go to the General Chemistry Office, room 1608 Gilman Hall.
Answer Key
2
1. (2 pts) Which of the following is a strong electrolyte? a) HCl(aq) b) H2O c) acetic acid d) ammonia e) all of the above are strong electrolytes 2. (6 pts) Consider an aqueous solution of acetic acid, write the formulas of all species expected to be
present (other than H2O). Circle the substance present at the highest concentration in the solution.
C2H3O–(aq) C2H3O2H(aq) H+(aq) or H3O+(aq) 3. (4 pts) What is the oxidation state of the sulfur atom in the metabisulfite ion S2O5
2–?
2x + 5 (–2) = –2 2x – 10 = –2 2x = +8 x = +4
4. (4 pts) Write the reaction equation that represents what happens when ammonia NH3 is dissolved in water. [Hint: what type of aqueous solution results from the dissolution of NH3?] For full credit, indicate the phases of all reactants and products (aq, l, s, g).
NH3(aq) + H2O(l) NH4OH(aq) or
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
5. (8 pts) How many liters of 0.23 molar NaOH is needed to neutralize 2.63 liters of 0.49 molar EDTA.
EDTA is a weak acid that can donate four protons H+ and is symbolized in the following equation as H4Y. For full credit, show all work, use units, and report your answer to the correct number of significant figures.
4 NaOH(aq) + H4Y 4 H2O(l) + Na4Y(aq)
2.63 L H4Y ⎝⎜⎛
⎠⎟⎞0.49 mol H4Y
1 L ⎝⎜⎛
⎠⎟⎞4 mol NaOH
1 mol H4Y ⎝⎜⎛
⎠⎟⎞1 L
0.23 mol NaOH = 22 L (2 sig figs)
3 6. (14 pts) In 1847, the Italian scientist Ascanio Sobrero first synthesized nitroglycerin, the explosive
agent in dynamite. Nitroglycerin C3H5(NO3)3 is produced by the reaction of glycerin C3H5(OH)3 and nitric acid:
C3H5(OH)3 + 3 HNO3 C3H5(NO3)3 + 3 H2O
What is the theoretical yield in grams of nitroglycerin that can be formed
when 251 g of glycerin C3H5(OH)3 is added to 2.00 liters of 6.00 Molar HNO3(aq)? For full credit, show all work, use units, and report your final answer to the correct number of significant figures.
251 g glycerin ⎝⎜⎛
⎠⎟⎞1 mol glycerin
92.1 g ⎝⎜⎛
⎠⎟⎞1 mol nitroglycerin
1 mol glycerin ⎝⎜⎛
⎠⎟⎞227 g
1 mol nitroglycerin = 619 g (3 sig figs)
2.0 L ⎝⎜⎛
⎠⎟⎞6.00 mol HNO3
1 L ⎝⎜⎛
⎠⎟⎞1 mol nitroglycerin
3 mol HNO3 ⎝⎜⎛
⎠⎟⎞227 g
1 mol nitroglycerin = 908 g
substance molar mass g/mol
C3H5(OH)3 glycerin 92.1
HNO3 nitric acid 63.0
C3H5(NO3)3 nitroglycerin 227
H2O water 18.0
Limiting reactant:
glycerin
theoretical yield of nitroglycerin
619 g (3 sig figs) grams
4 7. Nitroglycerin is a shock sensitive explosive and decomposes according to the following reaction:
4 C3H5(NO3)3 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) ΔH = –5811 kJ
A. (4 pts) This reaction is ___________ . a) endothermic b) exothermic c) neither exothermic nor endothermic B. (2 pts) This reaction is ___________ reaction. a) a redox or oxidation–reduction b) metathesis c) a neutralization d) a combination e) none of the above C. (4 pts) When nitroglycerin detonates under typical room temperature and pressure conditions,
________________ . a) work is done by the reaction on the surroundings b) work is done on the reaction by the surroundings c) no work occurs D. (2 pts) For the decomposition of nitroglycerin, the magnitude of ΔH is ____ than that of ΔE. a) greater than b) less than c) equal to E. (4 pts) Briefly but completely explain your answer to the last question. ΔE = q + w; q is negative (exothermic) and w is negative (work done on the surroundings). The
reaction transfers heat energy to the surroundings and expends energy in doing work on the surroundings. Therefore the magnitude of ΔE is greater than ΔH. Both heat and work transfer for the reaction have energy transferred by the reaction on the surroundings
F. (8 pts) Calculate the heat transferred (q) when 454 g of nitroglycerin decomposes. Report your
answer in kilojoules kJ. For full credit show all work, use units, and report your answer to the correct number of significant figures. Reporting your answer with the correct sign will be awarded 1 extra credit point.
454 g ⎝⎜⎛
⎠⎟⎞1 mol nitroglycerin
227 g ⎝⎜⎛
⎠⎟⎞–5811 kJ
4 mol nitroglycerin = –2.91 × 104 kJ (3 sig figs)
answer: –2.91 × 104 kJ (3 sig figs)
kJ
5 8. (8 pts) Calculate the molar concentration of potassium ions K+ in a solution formed from mixing 10.0 mL of 0.10 Molar K3PO4(aq) with 10.0 mL of 0.10 Molar K2SO4(aq). Assume that the volumes
are additive. For full credit, show all work, use units, and report your answer to the correct number of significant figures. Please report your final answer in the box.
moles of K+ in 10.0 mL of 0.10 Molar K3PO4:
10.0 mL ⎝⎜⎛
⎠⎟⎞0.10 mol K3PO4
1000 mL ⎝⎜⎛
⎠⎟⎞3 mol K+
1 mol K3PO4 = 0.00300 mol K+
moles of K+ in 10.0 mL of 0.10 Molar K2SO4:
10.0 mL ⎝⎜⎛
⎠⎟⎞0.10 mol K2SO4
1000 mL ⎝⎜⎛
⎠⎟⎞2 mol K+
1 mol K2SO4 = 0.00200 mol K+
total volume in liters: 10.0 mL + 10.0 mL = 20.0 mL 20.0 mL ⎝⎜⎛
⎠⎟⎞1 L
1000 mL = 0.0200 L
molarity of K+: 0.00300 mol + 0.00200 mol0.0200 L = 0.00500 mol
0.0200 L = 0.25 Molar (2 sig figs) 9. (6 pts) When 1.00 g of H2 reacts with 1.00 g of O2 to form water, O2 is the limiting reactant. What is the
mass amount of excess H2 in grams? For full credit, show all work, use units, and report your answer to the correct number of significant figures. 2 H2(g) + O2(g) 2 H2O(l)
Amount of H2 that reacts: 1.0 g O2 ⎝⎜⎛
⎠⎟⎞1 mol O2
32.0 g ⎝⎜⎛
⎠⎟⎞2 mol H2
1 mol O2 ⎝⎜⎛
⎠⎟⎞2.02 g
1 mol H2 = .126 g H2
Excess H2: 1.00 g – 0.126 g = 0.87 g H2 (2 sig figs)
[K+(aq)] = 0.25 Molar (2 sig figs)
answer: 0.87 g H2 (2 sig figs) grams
6 10. (12 pts) Complete and balance the following reaction equations. For full credit, include the phase of all
reacting substances (s, l, g, or aq). If no reaction is expected to occur, write, “no reaction.” A. (NH4)2S(aq) + 2 AgNO3(aq) Ag2S(s) + 2 NH4NO3(aq) B. HCl(aq) + NaNO3(aq) no reaction C. H2SO4(aq) + 2 KOH(aq) K2SO4(aq) + 2 H2O(l) 11. (4 pts) Write the net ionic equation for the following reaction. For full credit, include the phase of all
substances (s, l, g, or aq). 2 HClO3 (aq) + Ba(OH)2 (aq) Ba(ClO3)2 (aq) + 2 H2O(l) H+(aq) + OH–(aq) H2O(l)
7 12. (8 pts) When 50.0 mL of 1.00 Molar HNO3 and 50.0 mL of 1.00 Molar NaOH are combined in a
constant pressure calorimeter, the temperature increases by 6.69oC. The total solution mass is 100.0 g and you may assume that the solution has the same specific heat as water, 4.18 J/g-oC.
What is the enthalpy change of the reaction: HNO3(aq) + NaOH(aq) H2O(l) + NaNO3(aq)? Note that the moles of HNO3 and NaOH that are combined are identical and stoichiometric; exactly
enough HNO3 is added to react with the NaOH. 50.0 mL of a 1.00 Molar solution contains 0.0500 moles of solute. For full credit, show all work, and report your answer to the correct number of significant figures.
qsolution = 100.0 g × 4.18 Jg oC × 6.69oC = 2.796 × 103 J
qrxn = –2.796 × 103 J
heat in kJ: –2.796 × 103 J ⎝⎜⎛
⎠⎟⎞1 kJ
1000 J = –2.796 kJ
ΔHrxn = 1 mol HNO3 or NaOH ⎝⎜⎛
⎠⎟⎞–2.796 kJ
0.0500 mol = –55.9 kJ (3 sig figs) 13. (4 pts) Using the following information 2 CO(g) 2 C(s) + O2(g) ΔH = +222 kJ What is the enthalpy of reaction of the following reaction?
C(s) + 12 O2(g) CO(g) ΔH = ?
For full credit show all work, use units, and report your answer to the correct number of significant
figures. 2 C(s) + O2(g) 2 CO(g) ΔH = –222 kJ
C(s) + 12 O2(g) CO(g) ΔH =
–222 kJ2 = –111 kJ
answer:
–55.9 kJ (3 sig figs) kJ
answer: –111 kJ (3 sig figs) kJ
8 Chem 177 Periodic Table Exam II
9 1 cal = 4.184 joules
specific heat = sp. ht. = q
mass in g × ΔT
q = sp.ht. × m × ΔT = cs × m × ΔT ΔE = – CvΔT density of water at 25oC: 1.00 g/mL
specific heat of water: 4.18 J
g oC
Substance Molar Mass H2 2.02 g/mol O2 32.0 g/mol
H2O 18.0 g/mol
top related