This document provides sample chemistry problems and questions related to topics like: - Writing balanced equations for neutralization reactions - Defining terms like oxidation, reduction, concentration, and indicators - Determining oxidation states of elements in compounds - Identifying redox, precipitation, and acid-base reactions - Performing calculations involving molarity, moles, and concentration The problems cover a wide range of general chemistry concepts.

1. KEY
GENERAL CHEMISTRY-I (1411)
S.I. # 17
1. Write the balanced molecular AND net ionic equations for each of the following
neutralization reactions:
a. Aqueous acetic acid neutralized by aqueous potassium hydroxide
HC2H3O2 (aq) + KOH (aq)  KC2H3O2 (aq) + H2O (l)
HC2H3O2 (aq) + OH- (aq)  C2H3O2- (aq) + H2O (l)
b. Solid chromium(III) hydroxide reacting with nitric acid
Cr(OH)3 (s) + 3HNO3 (aq)  Cr(NO3)3 (aq) + 3H2O (l)
Cr(OH)3 (s) + 3 H+ (aq)  3H2O (l) + Cr3+ (aq)
c. Aqueous hypochlorous acid and aqueous calcium hydroxide react
Ca(OH)2 (aq) + 2HClO (aq)  Ca(ClO)2 (aq) + 2H2O (l)
HClO (aq) + OH- (aq)  ClO- (aq) + H2O (l)
2. Define the following:
a. Oxidation: is the loss of electrons by a substance
b. Reduction: is the gain of electrons by a substance
c. Concentration: amount of substance in terms of molarity
d. Molarity: the number of moles of solute per liters of solvent
e. Indicator: can be used to show the end point of the titration
3. Can oxidation occur without accompanying reduction?
Oxidation and reduction can only occur together, not separately. When a metal reacts
with oxygen, the metal atoms lose electrons, and the oxygen atoms gain electrons. Free
electrons do not exist under normal conditions. If electrons are lost by one substance,
they must be gained by another and vice versa.
4. Determine the oxidation number for the indicated element in each of the
following compounds:
a. Ti in TiO2 +4
-
b. Sn in SnCl3 +2
c. C in C2O42- +3
d. N in N2H4 -2
e. N in HNO2 +3
2-
f. Cr in Cr2O7 +6
5. Which of the following are re-dox reactions? For those that are, indicate which
element is oxidized and which is reduced. For those that are not, indicate whether
they are precipitation or acid base reactions.
a. Cu(OH)2 (s) + 2HNO3 (aq)  Cu(NO3)2 (aq) + 2H2O (l)
acid base reaction

2. KEY
b. Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)
oxidation – reduction reaction; Fe is reduced, C is oxidized
c. Sr(NO3)2 (aq) + H2SO4 (aq)  SrSO4 (s) + 2HNO3 (aq)
precipitation reaction
d. 4 Zn (s) + 10 H+ (aq) + 2 NO3- (aq)  4 Zn2+ (aq) + N2O (g) + 5 H2O (l)
oxidation – reduction reaction; Zn is oxidized, N is reduced
6. a. Calculate the molarity of a solution made by dissolving 0.145 mol Na2SO4 in
enough water to form exactly 750 mL of solution. b. How many moles of KMnO4
are present in 125 mL of a 0.0850 M solution? c. How many milliliters of 11.6 M
HCl solution are needed to obtain 0.255 moles of HCl?
a. M = mol solute; 0.145 mol Na2SO4 = 0.193 M Na2SO4
L solution .750 L
b. mol = M x L; 0.0850 mol KMnO4 x 0.125 L = 1.06x10-2 mol KMnO4
1L 1L
c. L = mol; 0.255 mol HCl = 2.20x10-2 L or 22.0 mL
M 11.6 mol HCl/L
8. a. How many grams are present in 50.0 mL of 0.360 M K2Cr2O7? b. If 4.28 g of
(NH4)2SO4 is dissolved in enough water to form 300 mL of solution, what is the
molarity of the solution? c. How many mL of 0.240 M CuSO4 contain 2.25 g of
solute?
a. 50.0 mL x 1 L x 0.360 mol K2Cr2O7 x 294.2 g K2Cr2O7 = 5.30 g K2Cr2O7
1000 mL 1 mol
b. 4.28 g (NH4)2SO4 x 1 mol (NH4)2SO4 x 1 x 1000 mL = 0.108 M (NH4)2SO4
132.2 g 300mL 1 L
c. 2.25 g CuSO4 x 1 mol CuSO4 x 1L x 1000mL = 58.7 mL solution
159.6 g 0.240mol 1L
9. Indicate the concentration of each ion present in the solution formed by mixing
a) 16.0 mL of 0.130 M HCl and 12.0 mL of 0.600 M HCl, b) 18.0 mL of 0.200 M
Na2SO4 and 15.0 mL of 0.150 M KCl solution.
a. H+ : 0.130 M x 16.0 mL + 0.600M x 12.0 mL = 0.331 M H+
28.0 mL
Cl-: concentration Cl- = concentration H+ = 0.331M Cl-
b. Na+: 2(0.200 M x 18.0 mL) = 0.218 M; K+ : = 0.150 M x 15.0 mL = 0.0682 M
33.0 mL 33.0 mL
SO42- : 0.200 M x 18.0 Ml / 33.0mL = 0.109 M; Cl- : 0.150 M x 15.0 mL/33.0 mL = 0.0682 M