FE CHEMISTRY REVIEW
STEVE DANIEL
sdaniel@mines.edu
Balance in basic solution:
NH2OH + MnO2(s) → Mn2+ + NO(g)
1. Assign oxidation numbers
-1+1-2+1
+4 -2
+2
+2-2
NH2OH + MnO2(s) → Mn2+ + NO(g)
NOTE
+1 -2
+1 -2
-2 +1
2 H2O ↔ H3O + OH2. Balance oxidation number changes
+
-1+1-2+1
+4 -2
+2
+2-2
2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g)
(+3)
(-2)
3. Balance charge
2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g)
net zero
net 6+
2NH2OH + 3MnO2(s) → 3Mn + 2NO(g) + 6OH2+
net zero
net zero
4. Balance remaining atoms (O and H)
2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) + 6OHtotal 8 O atoms
total 8 O atoms
So nothing is needed!
5. Check last atoms (H)
2NH2OH + 3MnO2(s) → 3Mn2+ + 2NO(g) + 6OHtotal 6 H atoms
total 6 H atoms
ASSIGNING OXIDATION NUMBERS
RULES IN PRIORITY ORDER
1. (oxidation numbers) = charge
2. Group IA (Li,Na,K,etc) assign +1
3. Group 2A (Be,Mg,Ca,etc) assign +2
4. B, Al assign +3
5. Hydrogen assign +1
6. Oxygen assign -2
Balance in acidic solution:
Cu2S(s) + NO3- → Cu2+ + SO42- + NO2(g)
1. Assign oxidation numbers
+1 -2
+5 -2
+2
+6 -2
+4 -2
Cu2S(s) + NO3 → Cu + SO4 + NO2(g)
-
2+
2-
2. Balance oxidation number changes
+1 -2
+5 -2
+2
+6 -2
Cu2S(s) +10 NO3 → 2Cu +
-
(+10)
2+
SO42-
+4 -2
+ 10NO2(g)
(-1)
3. Balance charge
Cu2S(s) +10 NO3- → 2Cu2+ + SO42- + 10NO2(g)
net 10-
net 2+
Cu2S(s) +10 NO3- + 12 H3O+ → 2Cu2+ + SO42- + 10NO2(g)
net 2+
net 2+
4. Balance remaining atoms (O and H)
Cu2S(s) +10 NO3- + 12 H3O+ → 2Cu2+ + SO42- + 10NO2(g)
total 36 H atoms
total 0 H atoms
Cu2S(s) +10 NO3- + 12 H3O+ → 2Cu2+ + SO42- + 10NO2(g) +
18H2O
5. Check last atoms (O)
Cu2S(s) +10 NO3- + 12 H3O+ → 2Cu2+ + SO42- + 10NO2(g) +
18H2O
42 O atoms on each side
STOICHIOMETRY
1. A copper ore is 5.00% Cu2S. How many grams of Cu
metal can be produced from 10.0 g of this ore.
Cu2S(s) +10 NO3- + 12 H3O+ → 2Cu2+ + SO42- + 10NO2(g) +
18H2O
3Cu2+ + 2Fe → 3Cu(s) + 2Fe3+
10.0g ore x 5.00gCu2S x moleCu2S x 2mole Cu x 63.54 g Cu =
100g ore 159.0gCu2S moleCu2S mole Cu
= 0.400 g Cu
2. How many grams iron are required to process the 10.0g of
ore?
10.0g ore x 5.00gCu2S x moleCu2S x 2mole Cu2+ x 2 mole Fe
100g ore 159.0gCu2S moleCu2S
3mole Cu2+
x 55.847g Fe = 0.234 g Fe
mole Fe
3. How many liters NO2(g) at 30.0oC and 620.0torr are formed
when the 10.0 g of ore is reacted?
10.0g ore x 5.00g Cu2S x mole Cu2S x 10 mole NO2 x
100g ore
159.0g Cu2S mole Cu2S
x .08205Latm x 303.15K x 760.0torr = .959 L
moleNO2K 620.0torr atm
4. How many mLs of 3.00M HNO3 solution are required to
react with 10.0g of the ore?
HNO3 + H2O → H3O+ + NO3- strong acid
10.0g ore x 5.00g Cu2S x mole Cu2S x 12 mole H3O+ x
100g ore
159.0g Cu2S
mole Cu2S
mole HNO3 x 1000 ml HNO3 = 12.6 ml HNO3
mole H3O+
3.00 mole HNO3
IDEAL GAS LAW
PV = nRT
V = nRT/P
R = .08205Lat/moleK
CONCENTRATION UNITS
Molarity = M = moles solute/L solution
Molality = m = moles solute/kg solvent
Normality = N = equivalents solute/L solution
REDOX
# equivalents/mole = ox.# change per formula
2Cr2O72- + N2H5+ → 2 NO2(g) + 4Cr3+
(-6)
(+12)
So Na2Cr2O7 has 6 eq/mole and N2H5Cl has 12eq/mole here
ACID-BASE
# eq/mole = # H+ gained or lost per formula
NH3 + H3PO3 → NH4+ + HPO32NH3 has 1eq/mole and H3PO3 has 2eq/mole in this reaction
5. Titration of 25.00 mL of a Ba(OH)2 solution requires 15.25
mL of 2.00 N HCl solution. What is the normality of
Ba(OH)2 solution? What is its molarity?
Ba(OH)2 + 2 HCl → Ba2 + 2 H2O + 2Cl.01525L HCl
x 2.00 eq HCl x eq Ba(OH)2 =
.02500LBa(OH)2
L HCl
eq HCl
= 1.22 eq Ba(OH)2/LBa(OH)2 or 1.22N
1.22 eq Ba(OH)2 x mole Ba(OH)2 = 0.610M
L Ba(OH)2
2 eq Ba(OH)2
6. When 25.0 g Zn metal and 300 mL 3.00N AgNO3 are
reacted, how many grams Ag metal form?
Zn(s) + Ag+ → Ag + Zn2+
(+2)
(-1)
Zn has 2 eq/mole and Ag+ 1 eq/mole
25. 0 g Zn x (mole Zn/65.38g) x (2 eq/mole) = 0.765 eq Zn
( .300L AgNO3)x(3.00 eqAg+/L AgNO3) = 0.900 eq Ag+
So Zn is the limiting reactant
(0.765 eq Zn)x(eq Ag)(mole Ag)(107.88gAg) = 82.5 g Ag
eq Zn eq Ag
mole Ag
HESS’ LAW
Heats of formation of HI(g) and HCl(g) are 26.5 and -92.3
kJ/mole, respectively. Calculate the standard enthalpy change
for:
2 HI(g) + Cl2(g) → I2(g) + 2 HCl(g)
H2(g) + Cl2(g) →2 HCl(g) H1o = 2mole HCl(-92.3kJ/mole)
2 HI(g) → I2(g) + H2(g) H2o = -2 mole HI(+26.5kJ/mole)
2 HI(g) + Cl2(g) → I2(g) + 2 HCl(g)
Hrxno = H1o + H2o = -184.6-(53.0) = -237.6 kJ/mole rxn
Hrxno = Σnprod∆Hof,prod - = Σnreact∆Hof,react
ELECTROCHEM
1. For the voltaic cell:
Zn/Zn2+(.200M)//Fe3+(.500M),Fe2+(.100M)/Pt
a. Write anode, cathode, and cell reactions (Eo = + 0.761v
and -0.771v for Zn→Zn2+ and Fe2+→Fe3+, respectively).
Anode: Zn → Zn2+ + 2eCathode: Fe3+ + e- → Fe2+
Cell: Zn + 2 Fe3+ → Zn2+ + 2 Fe2+
b. Calculate Eo and E for the cell.
Eocell = +0.761 – (-0.771) = +1.532v.
Ecell = 1.532 - .0591 log[Zn2+][ Fe2+]2 =
n
[Fe3+]2
= 1.532 –(.0591/2)log(.100)2(.200)/(.500)2 = +1.594v.
c. If each electrode compartment contains 500mL of solution
and the Zn electrode weighs 5.00g, how long can the cell
operate at an average current of 2.00amp?
.500mole Fe3+ x eq Fe3+ x .500L = 0.250 eq Fe3+
L solution
moleFe3+
5.00gZn x mole Zn x 2 eq Zn = 0.153 eq Zn
65.4gZn moleZn
.153 eq x 96487 coul x second = 7.38 x 103 seconds
eq
2.00 coul
ACID-BASE EQUILIBRIA
1. Calculate the pH of 0.50M HF solution. Ka = 7.1 x 10-4.
HF + H2O ↔ H3O+ + F.50-x
x
x
-4
+
Ka = 7.1 x 10 = [H3O ][ F-]/[HF] = x2/(.50-x)
x = 1.85 x 10-2 pH = -log[H3O+] = 1.73
2. If 0.20 mole NaF is dissolved in 300.0mL 0.50M HF, what
is the solution pH?
HF + H2O ↔ H3O+ + F.50-x
x
(.20/.300)+x
7.1 x 10-4= [H3O+][ F-]/[HF] = x(.667+x)/(.50-x)
x = [H3O+] = 5.3 X 10-4
pH = 3.28