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CHEMICAL REACTIONS
1. CO(g) + 2 H2(g)  CH3OH(g)
(balanced)
2. B2O3(s) + 3 Mg(s)  2 B(s) + 3 MgO(s)
(balanced)
3. Ca(s) + 2 H2O(l)  H2(g) + Ca(OH)2(aq) (balanced)
4.
a) ___ FeCl3(aq) + _3_ KOH(aq)  ___ Fe(OH)3(s) + _3_ KCl(aq)
b) ___ Pb(C2H3O2)2(aq) + _2_ KI(aq)  ___ PbI2(s) + _2_ KC2H3O2(aq)
c) ___ P4O10(s) + _6_ H2O(l)  _4_ H3PO4(aq)
d) ___ Li2O(s) + ___ H2O(l)  _2_ LiOH(aq)
e) ___ MnO2(s) + ___ C(s)  ___ Mn(s) + ___ CO2(g)
BALANCED
f) _2_ Sb(s) + _3_ Cl2(g)  _2_ SbCl3(s)
g) ___ CH4(g) + ___ H2O(g)  ___ CO(g) + _3_ H2(g)
h) ___ FeS(s) + _2_ HCl(aq)  ___ FeCl2(aq) + ___ H2S(g)
5.
a) Ba(NO3)2
SOLUBLE
b) K2CO3
SOLUBLE
c) Na2SO4
SOLUBLE
d) Cu(OH)2
INSOLUBLE
e) N/A
f) (NH4)3PO4
SOLUBLE
g) N/A
h) PbSO4
6.
INSOLUBLE
a) NH4Cl(aq) + H2SO4(aq)  NO REACTION: all possible products are soluble
in water.
b) N/A
c) _2_ NH4Cl(aq) + ___ Pb(NO3)2(aq)  _2_ NH4NO3(aq) + ___ PbCl2(s)
d) ___ CuSO4(aq) + _2_ KOH(aq)  Cu(OH)2(s) + K2SO4(aq)
e) N/A
f) N/A
7.
a) ___ K2SO4(aq) + ___ Ca(NO3)2(aq)  _2_ KNO3(aq) + ___ CaSO4(s)
b) N/A
c) _2_ AgNO3(aq) + ___ CaI2(aq)  _2_ AgI(s) + ___ Ca(NO3)2(aq)
d) N/A
e) N/A
f) ___ BaBr2(aq) + ___ Pb(NO3)2(aq)  ___ Ba(NO3)2(aq) + ___ PbBr2 (s)
STOICIOMETRY
2 H2 + O2  2 H2O
1.
10.0 g H2 * 1 mol H2 = 5.0 mol H2 * 2 mol H2O = 5 mol H2O = XXX
2 g H2
2 mol H2
10.0 g O2 * 1 mol O2 = 0.313 mol O2 * 2 mol H2O = 0.626 mol H2O* 18 g H2O = 11.3 g H2O
32 g O2
1 mol O2
1 mol H2O
2.
b) is TRUE
3.
a) 2.21x10-4 mol CaCO3 * 100.09 g CaCO3 = 0.0221 g CaCO3
1 mol CaCO3
b) 2.75 mol He * 4.003 g He = 11.0 g He
1 mol He
4.
a) 25.0 g Mg * 1 mol Mg * 1 mol CuCl2 = 1.03 mol CuCl2
24.31 g Mg 1 mol Mg
b) 25.0 g AgNO3 * 1 mol AgNO3
169.91 g AgNO3
*
1 mol NiCl2 = 0.0736 mol NiCl2
2 mol AgNO3
5.
2.25 g Cl2 * 1 mol Cl2 * 1 mol I2 * 254 g I2 =
71 g Cl2 1 mol Cl2 1 mol I2
8.05 g I2
6.
The limiting reactant is the one that gets used up first in a chemical reaction. It is
determined by the reactant that yields the least amount of theoretical product.
To determine the limiting reactant: Convert the starting mass (given) of each reactant to
moles of product. The lowest number is from the limiting reactant.
7.
% YIELD = actual yield
x 100 = 1.279 x 100 = 94.60 %
theoretical yield
1.352
GAS LAWS
1.
a) 105.2 kPa = 1.038 atm
b) 75.2 cm Hg = 0.990 atm
c) 752 mm Hg = 0.989 atm
d) 767 torr = 1.01 atm
2. The pressure increases proportionally
3.
4.
a) V1 = 541 ml
V2 = ?
P1 = 1 atm
P2 = 699 torr = 0.920 atm
P1V1 = P2V2
(1)(541) = (0.920)(V2)
V1 = 525 ml
V2 = ?
T1 = 25oC = 298 K
T2 = 50oC = 323 K
V1 = V2
T1
T2
(525) = V2
298
323
5. 2.01 g He = 0.502 mol He
6.52 g He = 1.63 mol He
V1 = 12.0 L
V2 = ?
6.
a) P = 1.01 atm
PV = nRT
b) P = ?
V2 = 569 ml
V1 = V2
n1
n2
V=?
V2 = 588 ml
(12.0) = V2
(0.502) (1.63)
n = 0.00831 mol
(1.01)V = (0.00831)(0.0821)(298)
V = 602 mL = 0.602 L n = 0.00801 mol
PV = nRT
c) P = 0.998 atm
PV = nRT
7. 2.52 g P4 * 1mol P4
123.88 g P4
n = 0.122 mol
PV = nRT
*
V2 = 39.0 L
T = 25oC = 298 K
V = 0.20 L
T = 310 K
P(0.602) = (0.00801)(0.0821)(310)
P = 0.339 atm
V = 629 mL = 0.629 L n = ?
T = 35oC = 308 K
(0.998)(0.629) = n (0.0821)(308)
n = 0.0248 mol
6 mol H2 = 0.122 mol H2
1 mol P4
T = 25oC = 298 K
P = 753 mmHg = 0.991 atm
(0.991)V = (0.122)(0.0821)(298)
V = 3.01 L
SOLUTIONS
1. Ionic solutes dissolve by the process of dissociation. The separate ions are shielded from each
other by water molecules.
DISSOCIATION
ionic compounds
Na
Cl
Na+
2.
Cl-
M = n / V (in liters)
a)
b)
c)
d)
M = 0.50/0.250 = 2.0-M
M = 0.50/0.500 = 1.0-M
M = 0.50/0.750 = 0.67-M
M = 0.50/1.0 = 0.50-M
3. 5.15 I2  0.020 mol I2
V = 225 ml = 0.225 L
M = n/V = (0.20)/(0.225) = 0.89-M
4. When the volume of a given solution is doubled (by adding water), the new concentration is
half the original concentration.
5. V1 = 75 ml
M1 = 0.211 M
M1V1 = M2V2
V2 = 125 ml
M2 = ?
(75)(0.211) = (125)M2
M2 = 0.127-M
EQUILIBRIUM
1. Activation energy represents the energy required from the outside for the reaction to proceed.
For a collision to be successful the molecules must e lined up properly and have enough
energy (activation energy)
2. A catalyst is a substance that speeds up a reaction without being consumed by the reaction. A
catalyst lowers the activation energy of the reaction.
3. Equilibrium is reached when there are no more observable changes in the system. This is
indicated in a chemical reaction by a two-headed arrow.
4. a) Keq = [NCl3]2
[N2][Cl2]3
b) Keq = [HI]2
[H2][I2]
c) Keq = [N2H4]
[N2][H2]2
5. LeChatlier’s principle states that whatever stress you put on a reaction, the equilibrium will
shift to relieve the stress.
6.
a) no effect (UO2 is a solid)
b) no effect (Xe is not involved in the reaction)
c) shift left (the concentration of HF will initially go down)
d) shift right (to replace the lost H2O)
e) shift left (pressure will go down, favoring the side with fewer GAS molecules)
7. Because Keq is less than one, the reaction favors the reactants. There will NOT be a relatively
large amount of product. This is not a useful reaction to get product. The reaction can be
made more useful by changing the temperature and/or the pressure.
ACIDS/BASES
1. Conjugate pairs differ by ONLY a H+ ion. NH4+ -- NH3
2. The water forms a hydronium ion (H3O+). This ion and water are conjugate pairs.
3.
a) CORRECT
b) NOT CORRECT
HBr – Br -c) CORRECT
d) NOT CORRECT
HNO3 – NO3-
BrO- -- HBrO
NO2- -- HNO2
4.
The first substance in each case is more acidic. It has a higher [H+]
5.
As the [H+] increases, the pH decreases. They are inversely related.
6.
pH = - log [H+]
a) pH = 3.00
b) pH = 3.66
c) pH = 10.0
d) pH = 6.33
ACIDIC
ACIDIC
BASIC
ACIDIC
7. The substances actually present are H+ and NO3-. No HNO3 molecules are present because
nitric acid is a strong acid and completely ionizes.
8.
9.
a) [H+] = 0.00010 M
b) [H+] = 0.0050 M
Buffers resist a change in pH.
pH = 4
pH = 2.3
b/c they are STRONG acids