KEY KEY KEY KEY KEY KEY
Balance the following reactions and identify each as
acid-base,
2 VO (s) +
BaCl2(aq) +
redox
3 Fe2O3 (s) 
K2SO2(aq)
or
6 FeO (s) +
-->
4 H+(aq) +
S2O32- (aq) +
Mg(OH)2(s) +
HCl(aq)
double displacement.
BaSO2(s) +
Sb2O5(s) 
--> H2O(l) +
Mg(OH)2(s) + 2 HCl(aq)
V2O5(s)
2 KCl(aq)
redox
double displacement.
2 SbO+(aq) +
MgCl2(aq)
--> 2 H2O(l) +
2 H2SO3(aq) redox
acid-base
MgCl2(aq)
Challenge: balance the following in acid solution.
You may add H+ and H2O as needed. I provided a couple of prompts to help you start.
AsI3(s) + K2Cr2O7(aq) --> I2(s) + Cr3+(aq) + AsO43-(aq) + K+(aq)
Net ionic equation: (AsI3 is soluble)
Oxidation numbers: in green above
Balanced equation: See steps above.
Concentration.
You have a 6.000M solution of HCl. In order to make 500. mL of 1.20M HCl, what
volume of the 6M solution do you need. Pay attention to significant figures.
(0.500L)(1.20M) = (6.00M)(V) V = 0.100L
A titration for determining the concentration of toxic arsenic in a sample can be
carried out using copper sulfate solution. The arsenic, in the form of arsenate ion,
AsO43-, is precipitated by copper(II) ion as copper(II) arsenate.
Write a balanced net ionic equation for the reaction of copper(II) and arsenate
ion.
3 Cu2+ + 2 AsO33-  Cu3(AsO3)2 (s)
2 points
A 50g sample is dissolved in 200mls. and titrated with 0.01M CuSO4 solution. The
sample takes 2.025 mls to react completely. Calculate the number of moles of
arsenate in the sample.
50 g sample (impure). So what we know is the amount of copper and we can
calculate back from that.
2.025 mls Cu2+ | 0.01mmol Cu2+ | 2mmol AsO3 3- = 1.35 x 10-5 mol AsO33| 1 ml Cu 2+
| 3 mmol Cu 2+
For sig figs, only 1 figure in concentration, so 1 x 10-5 mol AsO33-
1.4g of lithium sulfide (MW=45.938g) is reacted with 2.0g of aluminum chloride
(MW= 133.35g). Determine which is the limiting reactant and calculate the
maximum number of grams of product which can be produced (theoretical yield).
Hint: start with a balanced chemical reaction
3 Li2S + 2 AlCl3  Al2S3 + 6 LiCl
1.4g Li2S | 1 mol Li2S | 1 mol Al2S3 = .0102 mol Al2S3
45.88g Li2S | 3 mol Li2S
2.0g AlCl3 | 1 mol AlCl3 | 1 mol Al2S3 = .0075 mol Al2S3
133.43g AlCl3 | 2 mol AlCl3
AlCl3 is the limiting reagent.
Theoretical yield is .0075 mol Al2S3| 150 g Al2S3| = 1.125 g Al2S3
| mol Al2S3
|
If you included the LiCl in the product mass, I checked your calculations
and allowed that too.
Using heats of formation, calculate the ∆H of reaction for
PbS(s) + PbSO4(s) --> 2 Pb(s) + 2 SO2(g)
-93.303 -914.20
2(-297.04)
= 2(-297.04) –[-93.303 + -914.20] = 413.43 kJ
Given the following ∆H values for reactions, calculate the ∆H value for the final
reaction.
NCl3(l) + 1/2 O2(g) --> NOCl (g) + Cl2(g)
∆H = -178.41 kJ
2 SO2(g) + O2(g) --> 2 SO3(g)
∆H = -199.63 kJ
SO2(g) + Cl2(g) --> SO2Cl2(l)
∆H = -97.09 kJ
NCl3(l) + SO3(g) --> SOCl2(l) + NOCl(g)
∆H = __-175.69 kJ__
TAKEHOME
In order to analyze table salt for iodide content, a solution of 20g of salt in 20mLs
of water is titrated with 0.001M KMnO4 solution using the reaction
10 I- + 2 MnO4- + 16 H+ --> 5 I2 + 2 Mn2+ + 8 H2O .
The sample takes 11.0 mLs of permanganate to react. Calculate the concentration
of I- in the 20 mLs of solution and the percent by mass of I- in the 20g of salt.
11.0 ml | .001mmol MnO4 | 10mmol I| 1 ml
|
= 2.75 x 10-3 M
| 2mmol MnO4| 20 ml soln
4 points
5.5 x 10-5molesI | 126.9gI | 100%
| mol I
= 3.49 x 10-2 % I
| 20g sample
4 points
The pretreatment step involves the zinc reacting with iodate ion to produce iodide
and Zn2+ ion. Potassium iodate is actually the compound added to iodized salt.
Write and balance the net ionic equation for the reaction of iodate ion with zinc
metal. Identify the reducing and oxidizing agents and determine the oxidation
states of the atoms.
6H+ + 3Zn + IO3- = I- + 3Zn2+ + 3 H20
Zn (0) ; IO3- I(+5) ; I- (-1); Zn2+ (2+)
Zn reducing agent,
IO3-
oxidizer
1 for H’s, 2 for eqn.
4 points
1 point
The steps:
Mg(s) + F2(g)  MgF2(s) ΔH = ΔHf(MgF2(s))
Mg(s)  Mg(g)
ΔH = ΔHsublimation
Mg(g)  Mg2+(g)
ΔH = IE + IE2
F2(g)  2F(g)
ΔH = DE
-
F(g)  F (g)
2+
ΔH = -EA
-
Mg (g) + 2 F (g)
ΔH = -U (lattice energy)
Set up an equation for relating ΔHf(MgF2(s)) to the other ΔH’s in the list using
Hess’s law and the idea that ΔHf = sum of ΔH’s around the other path in the
diagram (up , across and down). 5
ΔHf(MgF2(s)) = ΔHsublimation + (IE + IE2) + DE +2(-EA) + (-U)
Solve this equation for the lattice energy U. 2
-U = ΔHf(MgF2(s)) - ΔHsublimation – (IE + IE2) – DE -2(-EA)
Using values from the tables on the next page, calculate the value of U in kJ/mol 4
=-1122.33-150-736.71-1441.43-243+656=-3037.47 kJ/mol
Compare the value of U obtained with the other ΔH’s in the tables below or in the
table in the back of the text. How does the magnitude of U compare with the
other typical ΔH’s for reactions? 2
Huge
Explain why the various ΔH’s for the steps listed on the previous page are positive
or negative, according to the sign convention for heat. 3
ΔHsublimation , (IE + IE2) , DE all + because energy must be provided.
EA, U, ΔHf(MgF2(s)) all – heat given off.