20- According to the balanced equation
O3 (g) + 3 NO (g) 
3 NO2 (g)
a) How many molecules of NO2 (g) could be produced from 25 molecules of O3 (g) ?
Solution and explanation: To solve this problem, firstly we need to identify what we are given and
what we are asked to answer. Herein we identify that we must find (x) number of molecules of NO2
produced from 25 molecules of O3 (g). Secondly, we write the identified elements below the relevant
formulas in the balanced equation as follow:
O3 (g)
+
3 NO (g) 
3 NO2 (g)
25 molecules
x molecules
Thirdly, we use factor-label method (dimensional analysis) based on our knowledge that 1 mole of any
gas contains 6.02 x 1023 molecules.
x molecules NO2 = 25 molecules O3 x
(1 mol O3)
x
23
(6.02 x 10 molecules O3)
(3 mol NO2)
(1 mol O3)
x (6.02 x 1023 molecules)
(1 mol NO2)
from the given balanced equation
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
= 75 molecules
Answer= 75 molecules NO2
b) How many moles of O3 (g) would be necessary to react with 80.6 L of NO (g)?
O3 (g)
x moles
+
3 NO (g) 
3 NO2 (g)
80.6 L
This is Volume-Mole Stoichiometry problem
x moleO3 = 80.6 L NO x (1 mol NO) x (1 mole O3)
(22.4 L NO)
(3 mol NO)
= 1.1994 mole or 1.2 mole O3
from the given balanced equation
Answer= 1.2 mole O3
c) How many moles of NO (g) would be needed to react with 6.00 grams of O3 (g)?
O3 (g) +
6.00 (g)
3 NO (g) 
x moles
3 NO2 (g)
This is Mass-Mole Stoichiometry problem. The GFM for O3= 48 g in 1 mole (48g/mol)
x mol NO = 6 g O3 x (1 mol O3 ) x (3 mol NO)
(48 g O3)
(1 mol O3)
= 0.375 mol NO
from the given balanced equation
Answer= 0.375 mole NO
21-
According to the unbalanced equation
NiSO4 + AlCl3  Al2(SO4)3
a)
+ NiCl2
How many grams of aluminum sulfate could be formed from 0.537 moles of aluminum chloride?
To solve this Mole-Mass problem we must firstly balance the unbalanced equation, identify what we are
given and what we are asked to answer. Herein we identify that we must find (x) number of grams of
Al2(SO4)3 that will be produced from 0.537 moles of AlCl3. Secondly, we write the identified
elements below the relevant formulas in the balanced equation as follow:
 Al2(SO4)3
3 NiSO4 + 2 AlCl3
0.537 mol
+ 3 NiCl2
x grams
Thirdly, we find the GFM of Al2(SO4)3
[ 2 Al= 2x27=54 3 S= 3x32=96
12 O = 12x16= 192];
342 g/mol (342 grams Al2(SO4)3 = 1 mole Al2(SO4)3) to use it in the factor-label method (dimensional
analysis) based on our knowledge of chemistry as follow:
x g Al2(SO4)3 = (0.537 mol AlCl3) x
(1 mol Al2(SO4)3
(2 mol AlCl3)
x
(342 g Al2(SO4)3)
(1 mol Al2(SO4)3)
= 91.83 g Al2(SO4)3
from the balanced equation
from GFM
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
= 91.83 g Al2(SO4)3
Answer= 91.83 g Al2(SO4)3
b) How many grams of NiSO4 would react with 51.4 grams of AlCl3?
 Al2(SO4)3
3 NiSO4 + 2 AlCl3
x grams
+ 3 NiCl2
51.4 grams
First Calculate GFM of AlCl3 =133.5 g/mole and NiSO4=155 g/mole then use dimensional analysis.
x g NiSO4 = 51.4 g AlCl3 x
(1 mol AlCl3
(133.5 g AlCl3)
(3 mol NiSO4)
(2 mol AlCl3)
x
x
(155 g NiSO4)
(1 mol NiSO4)
=89.51g
NiSO4
from GFM
from the balanced equation
from GFM
Answer = 89.51 g NiSO4
22- Given the unbalanced equation
C8H18 (g) +
a)
O2

(g)
CO2
+
(g)
H2O (ᶩ)
How many moles of water can be produced if 6.325 grams of O2 is used up?
To solve this Mole-Mass problem we must firstly balance the unbalanced equation, identify what we are
given and what we are asked to answer. Herein we identify that we must find (x) number of moles of
H2O that can be produced from 6.325 moles of O2. Secondly, we write the identified elements below the
relevant formulas in the balanced equation as follow:
2 C8H18 (g) + 25 O2
(g)

16 CO2
6.325 g
(g)
+ 18 H2O (ᶩ)
x mol
Thirdly, we find the GFM of O2
[ 2 O= 2x16=32 g/mol ] to use it in the factor-label method
(dimensional analysis) based on our knowledge of chemistry as follow:
x mol H2O = (6.325 g O2) x
(1 mol O2
(32 g O2)
x
(18 mol H2O)
(25 mol O2)
= 0.142 mol H2O
from GFM
from the balanced equation
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
Answer= 0.142 mol H2O
b)
How many grams of CO2 are produced if 114 grams of C8H18 (g) are burned?
O2 (g) 
C8H18 (g) +
H2O (ᶩ)
CO2 (g) +
To solve this Mass-Mass Stoichiometry problem we must firstly balance the unbalanced equation,
identify what we are given and what we are asked to answer. Herein we identify that we must find (x)
number of grams of CO2that will be produced if 114 grams of C8H18 (g) are burned. Secondly, we write
the identified elements below the relevant formulas in the balanced equation as follow:
2 C8H18 (g) + 25 O2
(g)

16 CO2
114 grams
(g)
+ 18 H2O (ᶩ)
x grams
Thirdly, we find the GFM of C8H18
and
8C=8x12=96
18H= 1x18=18___
114g/mol
CO2
as follow:
1C= 1x12=12
2O= 2x16=32
44g/mol
Fourthly, we use factor-label method (dimensional analysis) based on our knowledge of chemistry as
follow:
x g CO2 = 114 g C8H18 x
(1 mol C8H18)
(114 g C8H18)
from GFM
x
(16 mol CO2)
(2 mol C8H18)
x
from the balanced equation
(44 g CO2)
(1 mol CO2)
= 352 g CO2
from GFM
Finally, we cross-out the cross-multiplying units to be left out with the required (final) unit and then do
the mathematical calculations to get the final answer.
Answer = 352 g CO2