Chemistry 112
Second Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
1. (12 points) A compound is determined to be 52.14% C, 13.13% H, and 34.73% O. What
is the empirical formula of this compound?
2A. (4 points) Consider the following reaction:
P4(s) + 6F2(g) 6 4PF3 (g)
If I start with reaction with 5 g of P4 and 5 g of F2, what is my limiting reactant?
Molar Masses: P4 = 4x30.97 =123.88;F2=2x19.00=38.00;PF3 = 30.97+3(19.00)=87.97
Yield if P4 is limiting
5 g P4 x (1 mole P4/123.88 g P4) x (4 mole PF3/1 mole P4) = .161 mole PF3
Yield if F2 is ;limiting
5 g F2 x (1 mole F2/38.0 g F2) x (4 mole PF3/6 mole F2) = .0877 mole PF3
F2 gives the smaller amount of product so it is the limiting reactant
B. (4 points) What is the theoretical yield for the above reaction?
.0877 mole PF3 x (87.97 g PF3/1 mole PF3) = 7.71g PF3
2C. (4 points) If my final yield is 5 g of PF3, what is my % yield for this reaction?
% yield = Actual/Theoretical x 100%; =(5g/7.71g) x 100% = 64.85%
2
3. I am going to dissolve 15 grams of ZnCl2 in 90 mls of water. If the final volume of this
solution is 100 mls:
A. (4 points) What is the molar concentration of Zn2+ in this solution?
Molar mass of ZnCl2 = 65.38+2(35.45) = 136.28 g/mol
Molarity = mole/liter of solution
moles ZnCl2 = 15 g/136.28 g/mol = .110 mole
100 ml of solution = .1 Liters of solution
Molarity ZnCl2 = .110 moles/,1 liters = 1.1M
ZnCl2 dissolves in the following reaction: ZnCl2(s) 61 Zn2+(aq) + 2 Cl-(aq)
So if [ZnCl2] = 1.1 moles ZnCl2/1 liter x (1 mole Zn2+/1 mole ZnCl2) then [Zn2+]=1.1M
B. (4 points) What is the molar concentration Cl- in this solution?
if [ZnCl2] = 1.1 moles ZnCl2/1 liter x (2 mole Cl-/1 mole ZnCl2) then [Cl-]=2.2M
C. (4 points) What is the molar concentration of H2O in this solution?
90 mls of water = 90 grams of water
Molar mass of water = 16.00 + 2(1.008) = 18.016g/mol
Moles water = 90g/18.016 = 4.996 moles
Molarity of water = 4.996 moles water/0.1 liter of solution = 49.96M
4. Predict the products and their physical forms for the following reactions:
A.(3 points) AgNO3 (aq) + NaCl(aq)
6AgCl(s) + NaNO3(aq)
B. (3 points) Pb(NO3)2 (aq) + H2SO4(aq)
6PbSO4(s) + 2HNO3(aq)
C. (3 points) Na2S(aq) + CaCl2(aq)
6CaS(s) + 2NaCl(aq)
D. (3 points) HNO3(aq) + LiOH(aq)
6H2O(l) + LiNO3(aq)
3
5A. (4 points) Write a balanced molecular equation for the reaction of nickel(II) nitrate with
sodium hydroxide.
Ni(NO3)2 (aq) + 2 NaOH(aq) 6Ni(OH)2(s) + 2NaNO3(aq)
B. (4 points) Write a balanced complete ionic equation for the above reaction.
Ni2+(aq) + 2 NO3-(aq) + 2 Na+ (aq) + 2 OH-(aq) 6 Ni(OH)2(s) + 2 Na+(aq) + 2 NO3-(aq)
C. (4 points) Write a balanced net ionic equation for the above reaction.
Ni2+(aq) + 2 OH-(aq) 6 Ni(OH)2(s)
6. (12 points)Determine the oxidation state of all atoms in the following compounds
NaCl
C2H4
Na=+1
Cl = -1
H=+1
C
0 = 2C + 4H
0 = 2C +4
-4=2C
C= -2
H2SO4
Na2Cr2O7
H=+1
O= -2
S
0=2H+S+4O
0=2+S -8
6=S
Na=+1
O= -2
Cr
0=2Na+2Cr+7O
0= 2+2Cr -14
12=2Cr
Cr= +6
CO2
O= -2
C=+4
4
7. Balance the following redox reactions under acidic conditions:
A.(5 points)
K(s) + Cl2(g) 6KCl(s)
This one does not balance nicely by ½ reactions, works better by inspection
2K(s) + Cl2(g) 62KCl(s)
B. (5 points) Cu(s) + NO3-(aq) 6Cu2+(aq) + NO(g)
Cu0 6Cu2+
NO3-6NO
0
2+
Cu 6Cu + 2e
NO3-6NO + 2H2O
+
4H + NO3-6NO + 2H2O
3e- + 4H+ + NO3-6NO + 2H2O
x3
X2
3Cu0 63Cu2+ + 6e- 6e- + 8H+ + 2NO3-62NO + 4H2O
Total
0
+
3Cu + 6e + 8H + 2NO3-63Cu2+ + 2NO + 4H2O +6e3Cu0 (s) + 8H+ (aq) + 2NO3-(aq) 63Cu2+ (aq) + 2NO(g) + 4H2O (l)
C. (5 points) I-(aq) + ClO-(aq) 6I3- (aq) + Cl-(aq)
I- 6I3ClO-6Cl3 I 6I3
ClO-6Cl- + H2O
+
3 I 6I3 + 2e
2H + ClO-6Cl- + H2O
2e- + 2H+ + ClO-6Cl- + H2O
Total
3 I- + 2e- + 2H+ + ClO- 6 Cl- + H2O + I3- + 2e3 I- (aq)+ 2H+ (aq) + ClO- (aq) 6 Cl- (aq) + H2O(l) + I3- (aq)
8 Balance the following reactions under basic conditions:
A. (5 points) MnO4-(aq) + S2-(aq) 6MnS(s) + S(s)(aq)
MnO4- + S2- 6MnS
S2-6S
2MnO4 + S 6MnS + 4H2O
S2-6S + 2e+
28H + MnO4 + S 6MnS + 4H2O
5 e- + 8H+ + MnO4- + S2- 6MnS + 4H2O
X2
X5
10 e- + 16H+ + 2MnO4- + 2S2- 62MnS + 8H2O
5S2-65S + 10eCombine
5S2- + 10 e- + 16H+ + 2MnO4- + 2S2- 62MnS + 8H2O +5S + 10e7S2-+ 16H+ + 2MnO4- 62MnS + 8H2O +5S
Add OH- to both sides
7S2-+ 16H+ + 16OH- + 2MnO4- 62MnS + 8H2O +5S + 16 OH7S2-+ 16H2O + 2MnO4- 62MnS + 8H2O +5S + 16 OH7S2-(aq)+ 8H2O(l) + 2MnO4-(aq) 62MnS(s)+5S(s) + 16 OH-(aq)
5
B (5 points) NO2-(aq) + Al(s) 6 NH3(g) + AlO2-(aq)
NO2 6NH3
Al6AlO2-NO2 6NH3 + 2H2O
2H2O + Al6AlO2-+
7H + NO2 6NH3 + 2H2O
2H2O + Al6AlO2- + 4H+
6e- + 7H+ + NO2- 6NH3 + 2H2O
2H2O + Al6AlO2- + 4H+ + 3eX1
X2
+
6e + 7H + NO2 6NH3 + 2H2O
4H2O + 2Al62AlO2- + 8H+ + 6e+
4H2O + 2Al + 6e + 7H + NO2 6NH3 + 2H2O +2AlO2- + 8H+ + 6e2H2O + 2Al + NO2- 6NH3 + 2AlO2- + 1H+
Adding OH2H2O + 2Al + NO2- +OH- 6NH3 + 2AlO2- + 1H+ +OH2H2O + 2Al + NO2- +OH- 6NH3 + 2AlO2- + H2O
H2O(l) + 2Al(s) + NO2-(aq) +OH- (aq)6NH3(aq) + 2AlO2-(aq)