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This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Given that you have 14.5 moles of N 2 , how many moles of H 2 are theoretically needed to produce 22.5 moles NH 3 according to the reaction
N 2 + 3 H 2 → 2 NH 3?
- 21.8 moles H 2
- 2.33 moles H 2
- 43.5 moles H 2
- 33.8 moles H 2 correct
- 15.0 moles H 2
- 78.8 moles H 2
- 489 moles H 2
Explanation: The chemical equation is balanced. The stoi- chiometry indicates that 1 mol N 2 reacts with 3 mol H 2 to produce 2 mol NH 3.
? mol H 2 = 22.5 mol NH 3 ×
3 mol H 2 2 mol NH 3 = 33.8 mol H 2
002 10.0 points Consider the reaction
2 NH 3 + CH 3 OH → products.
How much NH 3 is needed to react exactly with 21 grams of CH 3 OH?
- 1.3 g
- 20 g
- 711 g
- 11 g
- 79 g
- 22 g correct Explanation: mCH 3 OH = 21 g
? g NH 3 = 21 g CH 3 OH ×
1 mol CH 3 OH 32 g CH 3 OH
×
2 mol NH 3 1 mol CH 3 OH
×
17 g NH 3 1 mol NH 3 = 22 g NH 3
003 10.0 points Consider the reaction
CaCN 2 + 3 H 2 O → CaCO 3 + 2 NH 3.
How much CaCO 3 is produced if 47.5 moles NH 3 are produced?
- 2380 g correct
- 0.950 g
- 4750 g
- 9500 g
- 0.238 g
- 23.8 g Explanation: nNH 3 = 47.5 mol
? g CaCO 3 = 47.5 mol NH 3 ×
1 mol CaCO 3 2 mol NH 3
×
100 g CaCO 3 1 mol = 2380 g
004 10.0 points Consider the following reaction:
CaCN 2 + 3 H 2 O → CaCO 3 + 2 NH 3
105.0 g CaCN 2 and 78.0 g H 2 O are reacted. Assuming 100% efficiency, which reactant is in excess and how much is leftover? The molar mass of CaCN 2 is 80.11 g/mol. The molar mass of CaCO 3 is 100.09 g/mol.
- H 2 O; 7.20 g left over correct
- H 2 O; 10.7 g left over
- CaCN 2 ; 70.8 g left over
- CaCN 2 ; 7.20 g left over
- CaCN 2 ; 10.7 g left over
- H 2 O; 70.8 g left over
Explanation:
CaCN 2 + 3 H 2 O → CaCO 3 + 2 NH 3
According to the stoichiometry 1 mol of CaCN 2 reacts with 3 moles of H 2 O. First we calculate the moles of CaCN 2 and of H 2 O as shown below.
? mol CaCN 2 = 105.0 g NaCN 2
×
1 mol CaCN 2 80 .11 g CaCN 2 = 1.31 mol CaCN 2
? molH 2 O = 78.0 g H 2 O ×
1 mol H 2 O 18 .0152 g H 2 O = 4.33 mol H 2 O
Since the stoichiometric ratio CaCN 2 to H 2 O is 1 to 3, to react completely 1.31 mol CaCN 2 we will require to use:
3 × 1 .31 = 3.93 mol H 2 O
We calculated that we have 4.33 mol H 2 O. Therefore water is in excess:
- 33 − 3 .93 = 0.40 mol H 2 O
? excess H 2 O = 0.40 mol H 2 O
×
18 .0152 g H 2 O 1 mol H 2 O = 7.20 g H 2 O
005 (part 1 of 3) 10.0 points Zinc citrate (Zn 3 (C 6 H 5 O 7 ) 2 ) is an ingredient in toothpaste. It is synthesized by the reac- tion of zinc carbonate with citric acid. 3 ZnCO 3 (s) + 2 C 6 H 8 O 7 (aq) → Zn 3 (C 6 H 8 O 7 ) 2 (aq) + 3 H 2 O(ℓ) + 3 CO 2 (g) If there are 1 mol of ZnCO 3 and 1 mol of C 6 H 8 O 7 , which is the limiting reactant?
- ZnCO 3 correct
- C 6 H 8 O 7 Explanation: nZnCO 3 = 1 mol nC 6 H 8 O 7 = 1 mol limiting reactant =? mole ratio :
2 mol C 6 H 8 O 7 3 mol ZnCO 3
nC 6 H 8 O 7 = 1 mol ZnCO 3 ×
2 mol C 6 H 8 O 7 3 mol ZnCO 3 = 0.666667 mol C 6 H 8 O 7 Because only 0. 666667 mol C 6 H 8 O 7 is needed to react with 1 mol ZnCO 3 , the limit- ing reactant is ZnCO 3.
006 (part 2 of 3) 10.0 points If there are 4.5 mol of ZnCO 3 and 1.75 mol of C 6 H 8 O 7 , which is the reactant in excess?
- C 6 H 8 O 7
- ZnCO 3 correct Explanation: nZnCO 3 = 4.5 mol nC 6 H 8 O 7 = 1 mol reactant in excess =? mole ratio :
2 mol C 6 H 8 O 7 3 mol ZnCO 3 nC 6 H 8 O 7 = 4.5 mol ZnCO 3 ×
2 mol C 6 H 8 O 7 3 mol ZnCO 3 = 3 mol C 6 H 8 O 7
ordered as they go from solid, to liquid, then to gas phase. Since both fusion and vapor- ization are endothermic changes with positive ∆H, ∆H total must be positive.
012 10.0 points What is the change in entropy (∆Svap) for the vaporization of ethanol (∆Hvap = 38 .6 kJ · mol−^1 ) at its standard boiling tem- perature (78. 4 ◦C)?
- −492 J · mol−^1 · K−^1
- −110 J · mol−^1 · K−^1
- 110 J · mol−^1 · K−^1 correct
- 492 J · mol−^1 · K−^1
Explanation:
∆Svap =
∆Hvap T
=
38 , 600 J · mol−^1 351 .4 K
= 110 J · mol−^1 · K−^1
013 10.0 points Which of the following constants are needed to calculate the change in enthalpy (∆H) for heating a sample of liquid water at 10 oC to steam at 150 oC. I) the heat capacity of water II) the enthalpy of fusion of water III) the heat capacity of ice IV) the enthalpy of vaporization of water V) the heat capacity of steam
- I, II, III, IV and V
- I and V only
- I, II and III only
- I, II, IV and V only
- I and III only
- I, IV and V only correct
- II and III only
Explanation:
The heat capacity of water is necessary to calculate heat needed to raise the liquid wa- ter’s temperature from 10 oC to 100 oC. The enthalpy of vaporization of water is necessary to calculate the heat needed to convert 10 gram of water at 100 oC to 10 grams of steam at 100 oC. Finally, the heat capacity of steam is necessary to calculate heat needed to raise the steam’s temperature from 100 oC to 150 oC.
014 10.0 points The ∆H vap◦ of methane is 8.519 kJ · mol−^1 and its ∆S◦ vap is 85.58 J · mol−^1 · K−^1. What it the boiling point of methane?
- − 263. 9 ◦C
- 0.01 K
- 372.54 K
- − 10. 05 ◦C
- 10.05 K
- 99.54◦C
- 0.01◦C
- 99.54 K correct
Explanation:
015 10.0 points How much heat is required to heat 0.1 g of ice at − 30 ◦C to steam at 100◦C? Use the approximate values cice = 2 J/g ·◦^ C csteam = 2 J/g ·◦^ C cwater = 4 J/g ·◦^ C
∆Hvap = 2260 J/g ∆Hfus = 340 J/g for your calculations.
- 300 J
- 80 J
- 74 J
4. 193 J
5. 187 J
- 306 J correct
Explanation: Heating ice from − 3 ◦C to 0◦C,
∆H = m cice ∆T = 0.1(2)(30) = 6 J.
Melting ice,
∆H = m ∆Hfus = 0.1(340) = 34 J.
Heating water from 0◦C to 100◦C,
∆H = m cice ∆T = 0.1(4)(100) = 40 J.
Vaporizing water,
∆H = m ∆Hvap = 0.1(2260) = 226 J.
Adding these terms, ∆H = q = 306 J.
016 10.0 points Which do you think would have the higher va- por pressure: Ethanol (C 2 H 5 OH) or dimethyl ether (CH 3 OCH 3 )?
- they would be the same
- neither has a vapor pressure
- dimethyl ether correct
- ethanol
Explanation: Dimethyl ether will have the higher va- por pressure since it has smaller IMF. Both molecules are polar and have similar disper- sion forces, but ethanol has hydrogen bond- ing.
017 10.0 points Consider two empty containers A and B whose volumes are 10 mL and 20 mL, respec- tively. You now put 1 mL of liquid H 2 O into each container and adjust the temperature of each container to 20◦C. The gas pressure in container B is found to be 17 torr. The gas pressure in container A is
- between 17 and 34 torr.
- 34 torr.
- 8.5 torr.
- 17 torr. correct
- greater than 34 torr.
Explanation: T = 20◦C + 273 = 293 K
P = 17 torr ·
1 atm 760 torr
= 0.02237 atm
n = 1 mL H 2 O ·
1 g H 2 O 1 mL H 2 O
1 mol H 2 O 18 g H 2 O = 0.0555 mol H 2 O Every liquid shows a characteristic vapor pressure that varies only with temperature, and not with the volume of vapor present, as long as both liquid and vapor are present and at equilibrium. Ideal Gas Law relates the volume and pressure of a sample of gas:
P V = n R T
V =
n R T P
=
(0.0555 mol)(0. 082058 L mol·atm·K )(293 K) (0.02237 atm) = 59. 65 L
Thus there must still be liquid left in each container and the pressure will be the vapor pressure.
018 10.0 points At 25◦C and 0.5 atm, which has the highest vapor pressure in a closed 3 liter container?
- They all have the same vapor pressure. correct
- 2 liters of water
- 2.99 liters of water
- 2.5 liters of water
An unknown liquid has a vapor pressure of 88 mmHg at 45◦C and 39 mmHg at 25◦C. What is its heat of vaporization?
- 32 kJ/mol correct
- 32000 kJ/mol
- 2000 kJ/mol
- 2000 J/mol
Explanation: Use the Clausius-Clapeyron Equation. Here, the only thing we don’t know is ∆Hvap.
ln
P 2
P 1
∆Hvap R
T 1
T 2
∆Hvap =
R ln
P 2
P 1
T 1
T 2
= 32.0 kJ/mol
024 10.0 points What is the vapor pressure of carbon disulfide at its normal boiling point?
- 2.5 atm
- 0.5 atm
- 2.0 atm
- 1.2 atm
- 0.3 atm
- 1.0 atm correct
Explanation: At ANY liquid’s normal boiling point, the vapor pressure = 1.0 atm.
025 10.0 points Answer the following about CO 2 using the provided phase diagram. What answer choice best describes the tem- perature at the triple point and critical point for CO 2 (triple point temperature, critical point temperature)?
Carbon Dioxide
Pressure, bar
Temperature, K
- Not enough information
- 310 K, 220 K
- 310 K, 310 K
- 220 K, 310 K correct
- 220 K, 220 K
Explanation:
026 10.0 points How many phase transitions occur by fol- lowing the steps described below?
step 1: At 50 K going from 90 atm to 20 atm step 2: At 20 atm going from 50 K to 750 K step 3: At 750 K going from 20 atm to 55 atm
b
Pressure, atm
Temperature, K
- 0 phase boundaries are crossed
- 2 phase boundaries are crossed
- 3 phase boundaries are crossed
- 1 phase boundaries are crossed correct
Explanation:
027 10.0 points At 20◦C the vapor pressure of dry ice is 56. atm. If 10 g of dry ice (solid CO 2 ) is placed in an evacuated 0.25 L chamber at a con- stant temperature of 20◦C, will all of the solid sublime?
- Yes correct
- No
- Only if it is at the triple point
Explanation: The dry ice will sublime until the pressure is equal to the vapor pressure. There are 0. moles of CO 2. If all sublimes the pressure will be less than the vapor pressure, so the solid will never reach equilibrium with the gas, and the system will be all gas.
028 10.0 points The phase diagram for carbon dioxide is given below.
Solid
Liquid
Vapor
Temperature, K
Pressure, atm
If the triple point is at 5.1 atm and − 56 ◦^ C, at 1 atm and room temperature,
- gaseous carbon dioxide is the stable phase. correct
- solid carbon dioxide melts.
- liquid carbon dioxide is the stable phase.
- solid carbon dioxide is the stable phase.
- gaseous carbon dioxide condenses.
Explanation:
029 10.0 points Carbon Dioxide
A
B
Pressure, bar
Temperature, K
A sample of carbon dioxide is stored at 104 bar and 250 K. This sample is then decompressed to 10^0 bar at constant temper- ature. Then, at constant pressure it is heated
