1.)A hydrocarbon has the formula C12H20. 61.51 g of it are burned in the
presence of excess pure oxygen gas. How many grams of water are
produced in this combustion?
Use these atomic weights: 12.01(C), 1.008(H), and 16.00(O).
First, write the balanced equation.
C12H20 + 17 O2 -- 12 CO2 + 10 H2O
61.51 g C12H20 x
1 mol C12H20 x
164.28 g C12H20
10 mol H2O x
1 mol C12H20
18.016 g H2O
1 mol H2O
= 67.46 g H2O
Use the flowchart:
A is what you start with, B is what you want.
Quantity A - mol A - mol B - Quantity B
mass A - mol A - mol B - mass B
See how everything cancels?
2.)How many mL of water are needed to dilute 32.07 mL of a solution 1.53 M
in HCl to 0.154 M HCl?
Use C1 V1 = C2 V2
where C1 and C2 are concentrations and V1 and V2 are volumes.
C1 V1 = C2 V2
32.07 mL x 1.53 M = 0.154 M x V2
V2 = 318.6 mL
This represents the TOTAL volume.
The amount that needs to be ADDED is 318.6 mL -32.07 mL = 286.5 mL
3.): What is the oxidizing agent in the following REDOX reaction?
Zn(s) + Cu2+(aq) ===> Cu(s) + Zn2+(aq)
The oxidizing agent is the one that gets REDUCED.
Zn - Zn2+ (0 - 2) this is OXIDIZED (charge went up)
Cu2+ - Cu (2 - 0) this is REDUCED (charge went down)
Cu2+ is the oxidizing agent.
4.)The cyanide ion, CN-, has many of the properties of halide ions. On this
basis, you would expect it to form a precipitate with the Na+ cation
true or false?
False. All sodium salts of halides are soluble, and so is NaCN.
5.)The cyanide ion, CN-, has many of the properties of halide ions. On this
basis, you would expect it to form a precipitate with the Ag+ cation.
true or false?
True. AgCl, AgBr and AgI are all insoluble salts, so is AgCN.
6.)Using the following portion of the activity series for oxidation half
reactions, determine which reaction will occur:
K ===> K+ + eAl ===> Al3+ + 3eFe ===> Fe2+ + 2eSn ===> Sn2+ + 2e(Assume aqueous solution. I have left out the (s) and (aq) symbols.)
The general rule here is that any metal will react with a metal ION below it on the
activity table.
K will react with Al3+, Fe2+ and Sn2+
Al will react with Fe2+ and Sn2+
Fe will react with Sn2+
8.)In the balanced equation for the reaction where dichromate oxidizes aq.
HCl to chlorine gas, the number of waters is (acidic solution)...
Here is the equation we need to balance…
Cr2O72- + Cl- - Cl2 + Cr3+
Split into half reactions
Cr2O72- - Cr3+
Cl- - Cl2
Balance
Cr2O72- - 2 Cr3+
2 Cl- - Cl2
Balance O
Cr2O72- - 2 Cr3+ + 7 H2O
2 Cl- - Cl2
Balance H
14 H+ + Cr2O72- - 2 Cr3+ + 7 H2O
2 Cl- - Cl2
Balance electrons
6 e + 14 H+ + Cr2O72- - 2 Cr3+ + 7 H2O
2 Cl- - Cl2 + 2 e
Multiply to cancel electrons
6 e + 14 H+ + Cr2O72- - 2 Cr3+ + 7 H2O
6 Cl- - 3 Cl2 + 6 e
Add reactions
6 e + 14 H+ + Cr2O726 Cl14 H+ + Cr2O72- + 6 ClThe answer is 7 H2O
- 2 Cr3+ + 7 H2O
- 3 Cl2 + 6 e
- 2 Cr3+ + 7 H2O + 3 Cl2
9.)Which of the following compounds is an Arrhenius base?
CH3OH
CH3COOH
HOCl
KOH
KOH..it the only one that forms OH- when dissolved in water. This is the best
indication of an Arrhenius base.
KOH - K+ + OH-
10.)which of these are cations which are precipitated by the chloride anion.
Ag+
H+
Ba2+
Na+
NH4+
Ag+ is the only one…all others are soluble.
11.)How many sulfur atoms are in 35.0 g of aluminum sulfide?
Al2S3
Go from mass - moles - molecules - atoms
35.0 g Al2S3 x 1 mol Al2S3 x
150.12 g Al
23
= 4.21 x 10 atoms S
See how everything cancels?
6.022 x 1023 molecules x
1 mol Al2S3
3 atoms S
1 molecule Al2S3
13.)Which of the following compounds has the highest percentage of
carbon by mass?
CH3COOOH
C2H5OH
C2H6
CaC2
H2C2O4
Use the formula:
# atoms C/12.011 g x 100 % = % Carbon
MW of compound
They each have 2 C atoms, so the one with the smallest MW will have the
highest % C. (C2H6)
14.)How many moles of NaOH are in 22.0 mL of 0.150M NaOH?
0.150 mol
1L
x
1L
1000 mL
x 22.0 mL = 0.00330 mol
A conversion from L to mL is necessary here.
15.)The formula for zirconium phosphate is Zr3(PO4)4. On the basis of this
information, the formula for the sulfide of Zr would be expected to be
PO4 has the charge -3. Total negative charge is 4 x -3 = -12.
Total negative charge = Total positive charge
Total positive charge = +12. Therefore EACH Zr has a +4 charge (12/3).
S has a -2 charge. If Zr has a +4 charge.
ZrS2 will be the formula of the sulfide.
16.)Which of the following samples contains the largest number of atoms?
1.0 mol propane gas, C3H8
5.0 mol iodine vapor, I2
2.5 mol ammonia, NH3
10.0 mol neon gas, Ne
3.0 mol water
use # mole x 6.022 x1023 molecule x
1 mole
total atoms
1 molecule
Cutting out the constant values…
1.0 mol propane gas, C3H8
5.0 mol iodine vapor, I2
2.5 mol ammonia, NH3
10.0 mol neon gas, Ne
3.0 mol water
1.0 x 11
5.0 x 2
2.5 x 4
10.0 x 1
3.0 x 4 = 12 (answer)
17.)Write a balanced equation for the combustion of butane, C4H10. The
coefficient of oxygen in the balanced equation is.
C4H10 + O2 - CO2 + H2O
C4H10 + O2 - 4 CO2 + 5 H2O (balance C and H)
C4H10 + (13/2)O2 - 4 CO2 + 5 H2O (count up O on left and divide by 2 on right)
2 C4H10 + 13 O2 - 4 CO2 + 5 H2O
18.)Chlorine was passed over 1.10 g of heated titanium and 3.54 g of a
chloride of Ti were obtained. What is the empiricial formula of this
chloride?
Mass of chloride of Ti = 3.54 g, mass of Ti = 1.10 g therefore
Mass of Cl = Mass of chloride – mass of Ti = 2.54 g Cl
Mass of Ti = 1.10 g
Moles of Ti = 1.10 g/(47.88 g/mol) = 0.0230 mol
Mass of Cl = 2.44 g Moles of Cl = 2.44 g/(35.45 g/mol) = 0.0688 mol
0.0688 mol/0.0230 mol = about 3.
So formula = TiCl3
19.)Tartaric acid, which occurs in many fruits, has MW = 150.1 and contains
63.96% oxygen by mass. How many oxygen atoms are in each tartaric acid
molecule?
% O = # of atoms O/(16.00 g/mol) x 100%
MW TA
% O x MW TA
100 % x 16.00 g
=
# of atoms O
63.96 % x 150.1 g/mol = # of atoms O
100 % x 16.00 g/mol
# of atoms O = 6 atoms O
20.)formula weight of aluminum sulfate to the nearest 0.01 g/mol
Al2 (SO4)3 MW = 342.17 g/mol
21.)Which of the following contains the greatest mass of oxygen?
1.0 mol O2
1.0 mol ozone (O3)
1.0 mole potassium chlorate
0.50 mol potassium dichromate
0.50 mol Na2S2O8
use # mole x 6.022 x1023 molecule x
1 mole
atoms O
1 molecule
Cutting out the constant values…
1.0 mol O2
1.0 x 2
1.0 mol ozone (O3)
1.0 x 3
1.0 mole potassium chlorate (KClO3)
1.0 x 3
0.50 mol potassium dichromate (K2Cr2O7)
0.50 x 7
0.50 mol Na2S2O8
0.50 x 8 (answer)
22.)The analysis of an organic compound showed that it contains 0.175 mol
of C, 0.140 mol of H, and 0.0350 mol of N. Its molecular mass is about 160.
How many atoms of carbon are there in the empirical formula and the
molecular formula, respectively?
Dividing each of values by the smallest value (0.035), gives
C5H4N as the empirical formula.
The mass of the empirical formula is about 78 g/mol
Dividing this into 160 g/mol gives about 2 formula units, so the formula is
most likely: C5H4N x 2
C10H8N2 (answer is 10)
23.)The total number of atoms in one formula unit of ferric sulfate is?
Fe2(SO4)3
atoms 2 + 3 + 12 = 17 atoms
24.)An organic compound which has the empirical formula CHO has an
approximate molar mass of 145 g/mol. The best guess at its molecular
formula is
C2H2O2
C3H3O3
C4H4O4
C5H5O5
C6H6O6
Mass of CHO = 29 g/mol
Dividing this into 145 g/mol = 5…so C5H5O5 would be the best answer.
25.)250.0 mL of 3.00 M HCl are added to 400.0 mL of 6.00 M HCl. Assuming
that the volumes are additive, the final concentration is
Find moles in each sample first…
3.00 mol
1L
x
1L
1000 mL
x 250.0 mL = 0.750 mol
6.00 mol
1L
x
1L
1000 mL
x 400.0 mL = 2.40 mol
Total moles = 3.15 moles
Total volume = 650.0 mL (0.6500 L)
Concentration =
3.15 mol = 4.85 M
0.6500 L