Download Second exam 2014 with answers

City College, Chemistry Department
Chemistry 10301, sections H*, Prof. T. Lazaridis
Second Midterm exam, Nov 5th, 2014
Last Name: ____________________________________________
First Name: _____________________________________________
Note: There are 10 questions in this exam (check both sides of the sheet).
Fill in your answer in the blank space provided immediately following each
question. 1/2 point will be subtracted every time you report a numerical result with
an incorrect number of significant figures. The last sheet of this exam contains data
that may be needed to answer these questions. R= 0.08206 atm L/mol K.
1. (10) Using data in your datasheet, predict whether a reaction will take
place by mixing the following substances. If so, give the type of reaction and
write a net ionic equation for the reaction.
a. (5) BaS (aq) and CuSO4 (aq)
Precipitation.
Ba2+ (aq) + S2- (aq) + Cu2+ (aq) + SO42- (aq) ----> BaSO4 (s) + CuS (s)
b. (5) Cr(OH)3 (s) and HBr (aq)
Acid-base neutralization
Cr(OH)3 (s) + 3H+ (aq) ------> Cr3+ (aq) + 3 H2O (l)
2. (8) a) (5) Give the oxidation numbers of each element in the reactants and
products of the reaction CH4 + 2 NO2 ------> N2 + CO2 + 2 H2O .
b. (3) Identify the oxidizing and the reducing agent in this reaction.
a.
Reactants Products
C
-4
+4
N
+4
0
H
+1
+1
O
-2
-2
b. Oxidizing agent is NO2, reducing agent is CH4.
3. (12) A sample of NaCl weighing 1.4477 g is dissolved in water and
diluted to 250.0 mL. What volume of this solution will be needed to titrate
25.00 mL of 0.1000 M AgNO3 (aq)?
NaCl (aq) + AgNO3 (aq) ----> AgCl (s) + NaNO3 (aq)
Moles of AgNO3: 25.00 X10-3 L X 0.1000 M = 25.00 X 10-4
We need the same number of moles of NaCl.
We have 1.4477 g/58.442 (g/mol) = 0.024772 moles in 250.0 mL
25.00 X 10-4 moles X 250.0 mL/0.024772 mol = 25.23 mL
4. (10) A gaseous hydrocarbon has a density of 1.69 g/L at 24 oC and 743
torr. Combustion analysis revealed that its empirical formula is CH2. Find
the molecular formula of this hydrocarbon.
From the ideal gas law we find the molar mass:
1.69 g/L * 0.08206 L atm/mol K * (24+273) K
M=dRT/P = ---------------------------------------------------------- = 42.13 g/mol
743 torr X 1 atm/760 torr
The empirical formula mass is 14.026. 42.13/14.026= 3
So, molecular formula is C3H6.
5. (10) How many liters of CO2 (g) can be produced in the reaction of 5.24 L
CO (g) and 2.65 L O2 (g) if all three gases are measured at the same
temperature and pressure?
2 CO (g) + O2 (g) ------> 2 CO2 (g)
If all gases are measured at the same T,P, ratio of moles=ratio of volumes.
Find the limiting reactant:
5.24 L CO X (2molCO2/2molCO) = 5.24 L CO2 ß less
2.65 L O2 X (2molCO2/molO2) = 5.30 L CO2
So, CO is limiting. The actual amount produced will be 5.24 L.
6. (15) A 2.02 g sample of Al reacts with excess HCl (aq) and the liberated
hydrogen is collected over water at a temperature of 24 oC. What is the total
volume of the gas collected at a barometric pressure of 760.0 torr? (The
vapor pressure of water at 24oC is 22.4 torr). The reaction is:
2 Al (s) + 6 H+ (aq) -> 2Al3+ (aq) + 3 H2 (g)
First find the moles of H2 produced:
2.02 g Al/26.98 (g/mol) X (3/2) = 0.112 mol
The total pressure is 760.0 torr. The partial pressure of H2 is 760.0-22.4 torr.
The volume is found from the IG law:
V=nRT/P = 0.112 X 0.08206X(273+24) / (760-22.4) torr X 1 atm/760 torr=
= 2.82 L
7. (15) Calculate the standard enthalpy of the reaction
BrCl(g) --------> Br (g) + Cl (g)
using the following data:
Br2(l) -----> Br2 (g)
ΔHo = + 30.91 kJ
Br2(g) ------> 2Br (g)
ΔHo = +192.9 kJ
Cl2(g) ------> 2Cl (g)
ΔHo = +243.4 kJ
Br2(l) + Cl2(g) -----> 2 BrCl(g) ΔHo = + 29.2 kJ
Invert the last equation and multiply by ½. Multiply all others by ½ and
add.
ΔHo= -½*29.2 + ½*30.91 + ½*192.9 + ½* 243.4 = 219.0 kJ
8. (5) A 0.828-g sample of gasoline is burned in a bomb calorimeter with a
heat capacity of 9.89 kJ/K. The temperature in the calorimeter rises from
22.75 to 26.37 oC . Calculate the heat of combustion (ΔHo) of the gasoline,
in kJ per gram.
q= hc X ΔT = 9.89 kJ/K X (26.37-22.75) K = 35.8 kJ
heat of combustion: -35.8 kJ/0.828 g = -43.2 kJ/g
9. (5) Use data from your datasheet to determine the standard enthalpy
change at 25 oC for the reaction
NH4+ (aq) + OH- (aq) ------> NH3 (g) + H2O (l)
ΔHo= {enthalpy of formation of products}-{enthalpy of formation of reactants}=
ΔHof (NH3) + ΔHof (H2O(l)) – ΔHof (NH4+) – ΔHof(OH-) =
-46.11 +(-285.8) - (-132.5) - (-230.0) = 30.6 kJ
10. (10) How many liters of ethane, measured at 17oC and 714 torr, must be
burned to give off 2.75X104 kJ of heat?
2 C2H6 (g) + 7 O2 (g) ------> 4 CO2 (g) + 6 H2O (l)
ΔH=-3.12X103 kJ
Find moles of ethane: 2.75X104 kJ X 2 mol/ 3.12X103 kJ = 17.6 mol
Use IG law to find the volume:
V=nRT/P = 17.6 mol*0.08206 atmL/molK*(17+273) /714 torrX 1atm/760torr=446 L