CHEM 3: Midterm Exam 2: Fall 2014 (150 points)
Name:________________________________________
Lab Section: Tuesday
Wednesday
Thursday
INSTRUCTIONS: On the front AND back of the scantron, fill out your name and 7-digit student ID. Fill in
this information from left to right in the provided fields, and BUBBLE IN the values on the scantron. 1
point will be deducted from your exam grade for incomplete or erroneous filling of the scantron.
Please read each question carefully. Where the question specifies “MARK ALL THAT APPLY”, you may
select more than one answer on the scantron.
1.
MARK ALL THAT APPLY: Select ALL of the formulas below which are reasonable EMPIRICAL formulas.
A. CH3CO2H
B. (NH4)2SO4
C. C6H3(CH2Br)3
D. CH3CH2CCH
E. None of the above are reasonable empirical formulas.
2.
MARK ALL THAT APPLY: Select ALL of the formulas below which could be a possible MOLECULAR formula.
A. H2O2
B. K2O
C. C6H6
D. Mg3N2
E. None of the above are reasonable molecular formulas.
3.
The MOLECULAR formula of nitroglycerine is CH2CHCH2(ONO2)3. What is the corresponding EMPIRICAL formula for
nitroglycerine?
A. In this case the empirical formula is equal to the molecular formula: C3H5N3O9
B. CH2NO6
C. C3H5N3O6
D. CH2NO3
E. CH2NO2
4.
A substance has a molar mass of 236.72 g/mol with an EMPIRICAL formula of CCl3. What is the MOLECULAR formula of this
substance?
A. C2Cl2
B. CCl2
C. CCl6
D. C2Cl6
E. In this case, the empirical formula is equal to the molecular formula: CCl3
5.
An unknown substance is found to have an EMPIRICAL formula of CH2O. What is the exact MOLECULAR formula of this
substance?
A. C6H12O6
B. C12H22O11
C. C5O5H10
D. In this case, the empirical formula is equal to the molecular formula: CH 2O.
E. Cannot determine from information provided: a mass spectrometry experiment would have to be performed.
6.
A 160 gram sample of a hydrocarbon was found to contain 120 grams of carbon. What is the EMPIRICAL formula of this
hydrocarbon (CXHY)?
A. C12H16
B. C16H12
C. C3H4
D. CH4
E. CH
For questions # 7-13, categorize each reaction as:
A.) Synthesis
B.) Double Replacement
C.) Combustion
D.) Decomposition
There may be more than one of each category.
7.
3 Fe + 4 H2O → Fe3O4 + 8 H2
8.
SO3 + H2O → H2SO4
9.
3 Zn + 2 AlCl3 → 3 ZnCl2 + 2 Al
E.) Single Replacement.
10. Fe2(SO4)3 + 6 KOH → 3 K2SO4 + 2 Fe(OH)3
11. 4 Hg2CrO4 → 2 Cr2O3 + 8 Hg + 5 O2
12. 2 C20H42 + 61 O2 → 40 CO2 + 42 H2O
13. (NH4)2SO4 + CaCO3 → (NH4)2CO3 + CaSO4
14. When properly balanced, what should the coefficient on the water be in the equation below?
___ H3PO4 → ___ H4P2O7 + ___H2O
A. 1
B. 2
C. 3
D. 4
E. This equation is impossible to balance.
15. If the reaction below was balanced properly with the smallest whole number coefficients, what would be the stoichiometric
coefficient in front of the diatomic oxygen (O2)?
2 C10H20O + ___ O2 → 20 CO2 + 20 H2O
A.
B.
C.
D.
E.
29
30
39
40
This equation should actually be simplified to: C10H20O + 15 O2 → 10 CO2 + 10 H2O
16. If the reaction below was balanced properly with the smallest whole number coefficients, what would be the stoichiometric
coefficient in front of the water (H2O)?
6 H3BO3 → H4B6O11 + ___ H2O
A.
B.
C.
D.
E.
7
8
9
11
This equation is impossible to balance.
17. If the reaction below was balanced properly with the smallest whole number coefficients, what would be the stoichiometric
coefficient in front of the water (H2O)?
___ P4O10 + ___ H2O → ___ H3PO4
A.
B.
C.
D.
E.
2
4
6
8
10
18. If the reaction below was balanced properly with the smallest whole number coefficients, what would be the stoichiometric
coefficient in front of the diatomic oxygen (O2)?
___ FeS + ___ O2 → ___ Fe2O3 + ___ SO2
A.
B.
C.
D.
E.
2
4
6
7
8
19. If the reaction below was balanced properly, what would be the stoichiometric coefficient in front of the AgNO3?
___ AgNO3(aq) + ___ FeCl3(aq) → ___ Fe(NO3)3(aq) + ___ AgCl(s)
A.
B.
C.
D.
E.
1
2
3
6
9
20. MARK ALL THAT APPLY: Select ALL of the ionic compounds below that would form SOLUBLE, aqueous solutions in water.
A. Al2S3
B. PbSO4
C. MgCO3
D. K3PO4
E. None of the above would be soluble in water.
21. If Mn(NO3)4(aq) was mixed with (NH4)2CO3(aq), what is one possible product from this double replacement reaction?
A. Mn(CO3)2
B. MnCO3
C. (NH4)2(NO3)4
D. NH4(NO3)2
E. None of the above would be possible products from this mixing.
22. How many moles of magnesium oxide would be needed to produce 90 moles of oxygen gas (O2) according to the equation
below?
2 MgO → 2 Mg + O2
A. 270 moles
B. 180 moles
C. 90 moles
D. 45 moles
E. None of the above.
23. Considering the reaction below, how many moles of sodium metal (Na) would be needed to react with 45 moles of
nitrogen gas (N2)?
2 Na + 3 N2 → 2 NaN3
A. 60 moles
B. 45 moles
C. 30 moles
D. 15 moles
E. 10 moles
24. Assuming an excess of P4 is available in the reaction below, how many moles of tetraphosphorus hexaoxide could be
generated from 3.0 moles of diphosphorus tetroxide?
P4 + 6 P2O4 → 4 P4O6
A. 1.0 mole
B. 2.0 moles
C. 3.0 moles
D. 4.5 moles
E. 6 moles
25. Considering the reaction below, how many moles of NH3 could be produced from a mixture of 30 moles of H2 and 12 moles
of N2? Hint: calculate the theoretical yield of NH3 based on the limiting reagent.
N2 + 3 H2 → 2 NH3
A.
B.
C.
D.
E.
60 moles
30 moles
24 moles
20 moles
8 moles
26. The water-gas shift reaction shown below is an industrial process used to generate hydrogen gas. If 400 moles of carbon
monoxide reacts with 150 moles of water, how much of the reagent in excess would be left unreacted when the limiting
reagent is fully consumed?
A.
B.
C.
D.
E.
CO + H2O → CO2 + H2
250 moles of H2O in excess left unreacted
150 moles of H2O in excess left unreacted
150 moles of CO in excess left unreacted
250 moles of CO in excess left unreacted
Neither CO or H2O is in excess: the reactants have been mixed in a proper stoichiometric ratio.
27. What is the LIMITING reagent, if 5 moles of calcium hydroxide, 3 moles of tetraphosphorus decoxide, and 10 moles of
water are mixed together and react according to the equation below?
2 Ca(OH)2 + P4O10 + 2 H2O → 2 Ca(H2PO4)2
A.
B.
C.
D.
E.
Ca(OH)2
P4O10
H2O
Ca(H2PO4)2
There would be no limiting reagent starting from this mixture.
28. Syngas is produced from methane and water according to the reaction below. How many grams of methane would be
needed to fully react with 10.0 grams of water (H2O)?
CH4 + H2O → CO + 3 H2
A. 20.0 g
B. 11.3 g
C. 10.0 g
D. 8.90 g
E. Cannot determine from information provided.
29. How many grams of phosphorus pentachloride would be needed to generate 2.5 moles of POCl3 according to the reaction
below? Hint: be sure to report the answer to the correct number of significant figures; the molar masses have been provided
for convenience.
3 PCl5 +
Molar Masses (g/mol):
A.
B.
C.
D.
E.
1600 g
870 g
310 g
100 g
None of the above.
208.22
P2O5 →
141.94
5 POCl3
153.32
EXTRA CREDIT: (10 points) Show your work for credit; no partial credit will be awarded.
If 13.5 g of elemental sulfur (S8) react with 250 g of nitric acid (HNO3) according to the equation below, what is the maximum mass of
nitrogen dioxide (in grams) that could be obtained?
____ S8 + ____ HNO3 → ____ H2SO4 + ____ NO2 + ____ H2O
Start by balancing this equation: You can take the typical approach of balancing hydrogen first, balancing oxygen last.
Unfortunately, such a trial and error approach to this equation proves difficult (but not impossible).
As an alternative approach, we can set up a system of equations to solve for the stoichiometric coefficients. We can start by
assuming 1 mol of S8 reacts with X moles of HNO3, Y moles of H2SO4, Z moles of NO2 & W moles of H2O, where X, Y, Z, & W are
presently unknown variables.
1 S8 + X HNO3 → Y H2SO4 + Z NO2 + W H2O
Then, balance each element in terms of these variables:
On the LEFT
=
On the RIGHT
Sulfur:
8
=
Y
Hydrogen:
X
=
2Y + 2W
Nitrogen:
X
=
Oxygen:
3X
=
This relationship tells us if we have 1 S8 we must have 8 H2SO4
Since Y = 8, we can simplify this equation to X = 16 + 2W
Z
4Y + 2Z + W
Since X = Z & Y = 8, we can simplify this equation to: 3X = 32 + 2X + W
This leaves us with a system of 2 equations and two unknowns:
X = 16 + 2W
3X = 32 + 2X + W
Substitute 16 + 2W for X in the second equation: 3(16 + 2W) = 32 + 2(16 + 2W) + W and solve for W:
48 + 6W = 32 + 32 + 5W
W = 16
Then substitute W back into the expression for X: X = 16 + 2(16) = 48 = X. Since X = Z, this means Z also equals 48, and we’ve arrived
at the complete, balanced chemical equation:
1 S8 + 48 HNO3 → 8 H2SO4 + 48 NO2 + 16 H2O
Now use the stoichiometric coefficients as usual to complete the analysis and calculate the theoretical yield of NO 2: