Chemistry 112
Second Hour Exam
Name:____________
Please show all work for partial credit
1. (13 points) A substance that contains only C and H undergoes combustion analysis.
When 0.500 g of the substance is burned you get 1.69 g of CO2 and .346 g of H2O. What
is the empirical formula of this material?
2. (12 points) Consider the following reaction: P4(s) + 6F2(g) 64PF3(g). How many grams
of F2 are needed to produce 120. g of PF3 if the reaction has a 78.1% yield?
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3. (12 points) I am going to dissolve 15 grams of barium nitrate in 150 mls of water. What
is the molarities of barium nitrate, barium ion, and nitrate ion in this solution?
Barium nitrite = Ba(NO3)2
Moles Ba(NO3)2 = 15g/[137.3 + 2(14.01) + 6(16)] = 5.74x10-2
Molarity = Moles/Liters = 5.74x10-2/.15 l =.383 M Ba(NO2)2
1 Ba(NO2)2(s) 61 Ba+2(aq) + 2 NO2-(aq)
So [Ba+2] = .383 M
[NO2-] = 2(.383) = .765 M
4. (13 points) Predict the products, and give balanced complete ionic equations for the
following reactions. If there is no reaction, simply write N.R.
K2CrO4 and Fe(NO3)3
6 K+(aq) + 3CrO4-2(aq) + 2 Fe+3(aq) + 6NO3-(aq) 6Fe2(CrO4)3(s) + 6 K+(aq) + 6NO3-(aq)
H2CO3 and NaOH
2H+(aq) + CO3-2(aq) + 2Na+(aq) + 2OH-(aq) 62H2O(l) + 2Na+(aq) + CO3-2(aq)
Pb(NO3)2 and CsCl
Pb+2(aq) + 2NO3-(aq) + 2Cs+(aq) + 2Cl-(aq) 6PbCl2(s) + 2Cs+(aq)+ 2NO3-(aq)
NaOH and NH3
N.R. (A base doesn’t react with another base)
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5. (12 points) What volume of 0.5M H2SO4 is needed to neutralize 150 mls of 3M NaOH?
Molecular Reaction equation: H2SO4 (aq) + 2 NaOH (aq) 62H2O(l) + Na2SO4(aq)
Moles NaOH = Molarity x Volume(liters) = 3(.150) = .450 moles
.450 moles NaOH x (1 mole H2SO4/2 moles NaOH) = .225 moles H2SO4
To get .225 moles of H2SO4 we have M=moles/liters; .5=.225/X; X=.225/.5
=.450 liters or 450 mls
6A. (6 points) Assign the oxidation state for every element in the following reaction:
2PbS(s) + 3O2(g) 6 2PbO(s) + 2 SO2(g)
PbS: Pb = +2, S=-2
O2: O=0
PbO: Pb=+2, O=-2
SO2 S=+4, O=-2
6B. (3 points) In 6A which elements were undergoing oxidation, which are undergoing
reduction, and which elements and not undergoing any change in oxidation state?
S is getting oxidized because its oxidation number increases form -2 ro +4
O is getting reduced because its oxidation number decreases from 0 to -2
Pb in not changing its oxidation state
6C. (4 points) Write the oxidation and reduction ½ reactions for equation 6A.
Hard to make two separate reaction here, so let’s try this:
Oxidation of S
PbS 6 SO2
Reduction of O
O2 6PbO + SO2
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7. (10 points) Balance the following redox reaction under acidic conditions:
I-(aq) + ClO-(aq) 6I3-(aq) + Cl-(aq)
I-6I33I-6I33I-6I3- + 2e-
ClO-6ClClO-6Cl- + H2O
2H+ + ClO-6Cl- + H2O
2e + 2H+ + ClO-6Cl- + H2O
Combining:
3I- + 2e- + 2H+ + ClO-6I3- + 2e- + Cl- + H2O
Removing common terms:
3I-(aq) + 2H+ (aq) + ClO-(aq)6I3-(aq) + Cl-(aq) + H2O(l)
8. (15 points) Balance the following redox equation under basic conditions:
Cr(s) + CrO4-2 6Cr(OH)3
(Hint: Both ½ reactions include Cr(OH)3 )
Cr(s) 6Cr(OH)3
3H2O + Cr(s) 6Cr(OH)3
3H2O + Cr(s) 6Cr(OH)3 +3H+
3H2O + Cr(s) 6Cr(OH)3 +3H+ + 3e-
CrO4-2 6Cr(OH)3
CrO4-2 6Cr(OH)3 + H2O
5H+ + CrO4-2 6Cr(OH)3 + H2O
3e + 5H+ + CrO4-2 6Cr(OH)3 + H2O
Combining:
3H2O + Cr(s) + 3e- + 5H+ + CrO4-2 6Cr(OH)3 +3H+ + 3e- + Cr(OH)3 + H2O
Removing common terms:
2H2O + Cr(s)+ 2H+ + CrO4-2 62 Cr(OH)3
Removing H+ by adding OH2H2O + Cr(s)+ 2H+ + 2OH- + CrO4-2 62 Cr(OH)3 + 2 OH2H2O + Cr(s)+ 2H2O + CrO4-2 62 Cr(OH)3 + 2 OHFinal Equation:
4H2O(l) + Cr(s) + CrO4-2 (aq)62 Cr(OH)3(aq) + 2 OH-(aq)