Chemistry 112
Second Hour Exam
Name:____________
Please show all work for partial credit
1. I have an iron(III) chloride solution and a sodium hydroxide solution. Predict the chemical
reaction that occurs where these two reagents are mixed and write the corresponding
chemical equation:
A. (4 points) Molecular Equation:
FeCl3(aq) + 3NaOH(aq) 6 Fe(OH)3(s) + 3NaCl(aq)
B. (4 points) Complete Ionic Equation:
Fe3+(aq) + 3Cl-(aq) + 3 Na+(aq) + 3OH-(aq) 6 Fe(OH)3(s) + 3Na+(aq) + 3Cl-(aq)
C. (4 points) Net ionic equation
Fe3+(aq) + 3OH-(aq) 6 Fe(OH)3(s)
2. I dissolve 1 gram of silver nitrate (AgNO3) in 100 mLs of water, and 1 gram of potassium
chloride (KCl) in another 100 mLs of water. When I mix these two solutions together...
A. (3 points) Write a chemical equation for the reaction that occurs.
AgNO3(aq) + KCl(aq) 6 AgCl(s) + KNO3(aq)
This was the molecular equation, complete ionic or net ionic are also acceptable answers
B. (5 points) What is the limiting reagent in this reaction?
1g AgNO3 ×
1g KCl ×
1 mole AgNO3
1 mole AgCl ( product )
×
= .0059 mole
[107.9 + 14.01 + 3(16.00)]g AgNO3
1 mole AgNO3
1 mole KCl
1 mole AgCl
×
= .014 mole AgCl
[3910
. + 3545
. ]g KCl 1 mole KCl
AgNO3 yields the smallest amount of product so it must be the limiting reagent
C. (5 points) How much solid product will be formed by this reaction?
1g AgNO3 ×
1 mole AgNO3
1 mole AgCl
[107.9 + 35.45]g AgCl
×
×
= .084 g AgCl( s)
[107.9 + 14.01 + 3(16.00)]g AgNO3 1 mole AgNO3
1mole AgCl
2
3
3. (10 points) I have 5 L of 2.5 M HCl leftover from a lab. Before I can dispose of this acid,
I should neutralize it with sodium hydroxide (NaOH). I have a large quantity of 4M NaOH
available. How much 4M NaOH will be needed to neutralize my acid?
For neutralization the moles of acid = moles of base
Moles of acid = 5 R × 2.5 mole/R = 12.5 mole
Therefore I need 12.5 moles of base
How much 4M base do I need to get 12.5 moles of base?
12.5 moles = X R of base × 4 mole/R
Doing a little algebra
12.5 moles
= Xl = 3125
. l
4 moles / l
4. (12 points) Assign the oxidation states of the individual atoms in the following compounds:
A. NO3-(aq)
O= -2, N= +5
B. KMnO4(s)
O= -2, K = +1, Mn= +7
C. H2C2O4(aq)
O= -2, H = +1, C= +3
D. K 2Cr2O7(s)
O= -2, K = +1, Cr= +6
E. N2(g) N=0, N2 is the elemental form of Nitrogen so it is assigned a 0 oxidation number
4
5. (10 points) Balance the following REDOX equations. In the process of balancing the
equation, find and balance the two half-reaction equations and identify which half reaction is
the oxidation and which is the reduction.
Unbalanced equation: H+(aq) + Cr2O72-(aq) 6CO2(g) + Cr3+(aq) + H2O(l)
½ reaction involving C
CH46CO2
C is already balanced
Balance O with H2O:
2 H2O + CH46CO2
Balance H with H+:
2 H2O + CH46CO2 + 8H+
Balance charge with e2 H2O + CH46 CO2 + 8H++ 8 e Electrons on right, so this is an oxidation ½ reaction.
½ reaction involving Cr
Cr2O72- 6Cr+3
Balance Cr:
Cr2O72- 62Cr+3
Balance O with H2O:
Cr2O72- 62Cr+3 + 7H2O
Balance H with H+:
14 H+ + Cr2O72- 62Cr+3 + 7H2O
Balance charge with e6e - + 14 H+ + Cr2O72- 6 2Cr+3 + 7H2O
Electrons on left so this is a reduction ½ reaction.
Combining 2 equations, you need to multiply the first equation by 6 and the second equation by 8 so a total
of 48 electrons are generated in one reaction and consumed in the other reaction.
6x[2 H2O + CH46 CO2 + 8H++ 8 e-]
8x[6e- + 14 H+ + Cr2O72- 62Cr+3 + 7H2O]
+
12 H2O + 6 CH4 + 112 H+ + 8Cr2O72 + 48 e- 66CO2 + 48H+ + 16Cr+3 + 56H2O + 48 eremoving common terms:
6 CH4 + 64 H+ + 8Cr2O726 6CO2 + 16Cr+3 + 44H2O
Note: In the above equation all the coefficients are multiples of 2, and you should divide through by 2 to
remove this common term:
3 CH4 + 32 H+ + 4Cr2O726 3CO2 + 8Cr+3 + 22H2O
You should get the same answer if you multiplied the first equation by 3 and the second equation by 4.
5
6
6. (5 points) I have balanced the following redox equation under acid conditions:
8CN-(aq) + 4Ag(s) + 2O2(g) + 4H+(aq) 6 4 Ag(CN)2-(aq) + 2H2O(l)
Now balance the above equation under BASIC conditions.
8CN-(aq) + 4 Ag(s) + 2 O2(g) + 4H+ (aq) 6 4Ag(CN)2(aq) + 2H2O(l) Acidic
+ 4 OH+ 4OHOn this side H+ + OH-=H2O
On this side OH- remains unchanged
Net:
8CN-(aq) + 4 Ag(s) + 2 O2(g) + 4H2O (l) 6 4Ag(CN)2(aq) + 2H2O(l) + 4OH-(aq)
Removing the common water from both sides of the equation:
8CN-(aq) + 4 Ag(s) + 2 O2(g) + 2H2O (l) 6 4Ag(CN)2(aq) + 4OH-(aq)
Dividing both sides by 2 for lowest common denominator:
4CN -(aq) + 2 Ag(s) + O2(g) + H2O (l) 6 2Ag(CN)2(aq) + 2OH-(aq)
7. In the lab I have a piece of equipment called an HPLC that pumps liquids at 4000 PSI
A (5 points) what is this pressure in ATM
4000 lbs / in 2 ×
1 atm
= 272 atm
14.7 lbs / in 2
B. (5 points) What is this pressure in kPa?
4000 lbs / in 2 ×
1 atm
= 272 atm
14.7 lbs / in 2
8. I have my daughter’s balloon filled with 1 L of He at 1 ATM and 25oC.
A. (5 points) What is the volume of the balloon if I place it in a liquid N2 bath,
temperature -195oK?
PV=nRT
In this problem you change the temperature and watch the volume change, thus T and V are
variables, while nR and P are constants. Rearranging the ideal gas law so the variables are on one side and
the constants are on the other we have:
V/T = nR/P
And, since nR and P are constant, so they are the same before and after:
V1/T1=n1R1/P1=n2R2/P2 = V2/T2 so:
V1/T1 = V2/T2
V1 = 1R, T1 = 273.15 + 25 = 298.15K, T2 = 273.15-195 = 78.15K
1R/298.15K = XR/78.15K
7
X=1R(78.15K)/298.15K = .26R
B. (10 points) Assuming the balloon doesn’t explode, what would its volume be if I let
it go, and it ascended to 10 km (about 6 miles up). At 10 km the pressure is 210 torr, and the
temperature is -40oC.
Ideal Gas law: PV=nRT
In this problem you change the temperature and presure and then watch the volume change,
thus T, P and V are variables, while n and R are constants. Rearranging the ideal gas law so the variables
are on one side and the constants are on the other we have:
PV/T = nR
And, since nR are constant, so they are the same before and after:
V1 P1 /T1=n1R1=n2R2 = V2 P2 /T2 so:
V1 P1/T1 = V2 P2 /T2
V1 = 1R, P1 760 torr (1 atm) T1 = 273.15 + 25 = 298.15K,
V2 =X, P2=210 torr, T2 = 273.15-40 = 233.15K
(1R×760 torr)/298.15K = (XR ×210 torr)/233.15K
X=1R×(760/210)×(233.15/298.15) = 2.83R
9 A. (2 points) What is STP?
Standard pressure and temperature:
P=1 atm, T=273.15 K
B. (1 points) What is the volume of 1 mole of SF6 gas at STP?
22.42 R (all gases have this volume at STP)
10. (10 points) I have a chemical reaction that generates 2.5 mole of CO2 gas. What is the
volume of this gaseous product if the temperature is 35oC, and the pressure is 670 mm Hg?
Ideal Gas Law: PV=nRT
P = 670 mm Hg = 670/760 = .8816 Atm
V=?
n=2.5 mole
R=.08206 R@atm/K@mol
T=35 + 273.15 = 308.15K
.8816×? = .08206 (2.5) 308.15
?=(.08206 R@atm/K@mol×2.5 mol×308.15K)/.8816 atm
8
=71.7R