(a) Zn + NO3-1
Zn+2 + NO2-1
(acidic condition)
(b) Sn + ClO4-1
SnO2 + ClO3-1
(basic conditions)
1. (a) Calculate molarity if 25.0
mL of 1.75 M HCl diluted to 65.0 mL. (3)
(b) Calculate
molarity by dissolving 25.0g NaOH in 325 mL of solution. (3)
(c) Calculate
grams of solute needed to prepare 225 mL of 0.400 M KBr solution.
(3)
(d) Calculate
mL of 0.650M KNO3 needed to contain 25.0g KNO3.
(3)
(e) Which
are water soluble? Zn(NO3)2 AlCl3
AgBr FePO4 CuAc2 (3)
2. (a) Calculate grams Ca(OH)2
needed to completely neutralize 50.0 mL of 3.00 M HBr.
(4)
(b) Calculate
molarity of 35.0 mL KOH solution needed to completely neutralize 22.5 mL
of 1.75M H2SO4.
(4)
(c)
Calculate volume (mL) of 2.50M H2SO4 needed to completely
neutralize 10.0g NaOH(s). (4)
3. Write net ionic equation for the following:
(27)
(a) Zn(NO3)2(aq)
& Na2CO3(aq)
(b) PbAc2(aq) & KOH(aq)
(c) K2CO3(aq) & HBr(aq)
(d)
Cr(NO3)3(aq) & Mg(s)
(e) H2SO4(aq) & NaOH(aq)
(f) AgNO3(aq) & Zn(ClO3)2(aq)
(g) HBr(aq)
& KF(aq)
(h) Al(s) & Pb(NO3)2(aq)
(i) KCN(aq) & HCl(aq)
4. Assign oxidation numbers to underlined
atoms: (8)
(a) K2Cr2O7
(b) CaC2O4
(c) Al(NO2)3
(d) MnO4-1
5. Balance the following reactions:
(14)
(a) Zn
+ NO3-1
Zn+2 + NO2-1
(acidic condition)
(b) Sn
+ ClO4-1
SnO2 + ClO3-1
(basic conditions)
6. Use enthalpies of formation to calculate
DH:
(8)
(a) C2H5OH(l) C2H4(g)
+ H2O(l)
(b)
Fe2O3(s) + 3 Pb(s)
3PbO(s) + 2Fe(s)
7. (a) Calculate equilibrium temperature
(T) if 275g gold (s = .129 J/g/°C) at 85°C added to 50.0g water
at 22°C. (5)
(b) The addition
of 355 J to 25g of gold will change its temperature by how many °C?
(3)
(c) Calculate
DE
for reaction that is endothermic by 45.3 kJ while volume contracts by 25.0L
at 3.00 atm. (3)
(d) N2O
+ O N2O2
(DH = 125 kJ)
& N2O3 +
O N2O2
+ O2 (DH = 44 kJ)
(4)
Calculate
DH for N2O
+ O2 N2O3
Answers
1. (a) M1V1
= M2V2 (1.71
M)(25.0 mL) = M2(65.0 mL)
M2 = 0.658 M
(b) M = mol/L
= (25.0/40.0) / (0.325) = 1.92 mol/L
(c) g = (M)(L)(FW)
= (0.400)((0.225)(119) = 10.7 g
(d) (25.0g)(1
mol/101 g)(1000mL/0.650 mol) = 381 mL
(e) Zn(NO3)2
AlCl3 CuAc2
2. (a) 2 mol Ca(OH)2
= mol HBr
2 (g/74) = (3.00)(0.0500) 5.55
g Ca(OH)2
(b)
mol KOH = 2 mol H2SO4
( M)(mL) = 2( M)(mL)
( M)(35.0mL) = 2(1.75 M)(22.5 mL) 2.25M
KOH
(c)
2 mol H2SO4 = mol NaOH
2 ( M)(L) = g/FW
2 (2.50)(L) = 10.0/40.0 L = 0.0500 50.0
mL
3. (a) Zn+2+
CO3-2
ZnCO3(s)
(b) Pb+2 + 2OH-1
Pb(OH)2(s)
(c) 2H+1+
CO3-2
H2O(l) + CO2(g)
(d) 2Cr+3 + 3
Mg(s) 3Mgr+2
+ 2Cr(s)
(e) H+1
+ OH-1
H2O(l)
(f) NR
(g) H+1
+ F-1
HF(aq)
(h) 2Al + 3Pb+2
2Al+3 + 3Pb(s)
(i) H+1
+ CN-1
HCN(g)
4. (a) +6 (b) +3 (c) +3 (d) +7
5. (a) Ox:
Zn Zn+2
+
2e-1
Red: NO3-1 + 2H+1
+
2e-1
NO2-1 + H2O
Zn + 2H+ + NO3-1
Zn+2 + NO2-1 + H2O
(b) Ox:
4OH-1 + Sn
SnO2 + 2H2O
+
4e-1
Red: 2ClO4-1 + 2H2O
+ 4e-1 2ClO3-1
+ 4OH-1
2ClO4-1 + Sn
2ClO3-1 + SnO2
6. (a) DH
= -DH(C2H5OH) + DH(C2H4)
+ DH(H2O(l)) = -(-278kJ)
+ (52kJ) + (-286kJ) = +44kJ
(b) DH
= -DH(Fe2O3) + 3DH(PbO)
= -(-822kJ) + 3(-219kJ) = +165kJ
7. (a) heat gain water = heat loss
gold
mwswDT = -mgsgDT
(50.0)(4.18)(T-22) = (275)(.129)(85-T) T
= 31°C
(b) DT
= q/ms = (355J)/[(25.0g)(0.129J/g/°C)]
=
110°C
(c) DE
=
DH - PDV = 45.3kJ
- (3.00atm)(-25.0L)(0.101 kJ/atm-L) = 52.9 kJ
(d) DH
= 125 kJ + (-44 kJ) = 81 kJ